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Work, Energy and Power

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Chapter Notes

Overview
This chapter formalizes the physical meanings of work, energy, and power, moving beyond everyday language. It introduces the scalar (dot) product of vectors, defines work done by constant and variable forces, and establishes the work-energy theorem. The central idea is the conservation of mechanical energy for conservative forces, which is applied to gravitational potential energy, spring potential energy, and collisions. A student must understand that work is the transfer of energy, and that total mechanical energy is conserved only when non-conservative forces (like friction) do no work.
Key Concepts (12)
Scalar Product (Dot Product)
The scalar product of two vectors A and B, denoted as A.B, is defined as A.B = A B cos θ, where θ is the angle between the two vectors.
e.g. Calculating work done by a force F on a displacement d: W = F.d = F d cos θ.
Work
Work done by a constant force is the product of the component of the force in the direction of displacement and the magnitude of the displacement. W = (F cos θ) d = F.d.
e.g. A cyclist skidding to a stop: the frictional force of 200 N opposite to a displacement of 10 m does work W = 200 × 10 × cos 180° = -2000 J.
Kinetic Energy
Kinetic energy of an object of mass m moving with velocity v is K = (1/2) m v^2. It is a scalar quantity and measures the work an object can do by virtue of its motion.
e.g. A bullet of mass 50 g moving at 200 m/s has kinetic energy = 0.5 × 0.05 × 200^2 = 1000 J.
Work-Energy Theorem
The change in kinetic energy of a particle is equal to the work done on it by the net force. K_f - K_i = W.
e.g. A raindrop falling under gravity and resistive force: ΔK = 1.25 J, work by gravity = 10 J, so work by resistive force = -8.75 J.
Variable Force
A force whose magnitude and/or direction changes with position or time. Work done by a variable force in one dimension is given by the integral W = ∫_{x_i}^{x_f} F(x) dx.
e.g. A woman pushing a trunk with force reducing linearly from 100 N to 50 N over 20 m: work done = area under F-x graph = 1750 J.
Potential Energy
Potential energy is the stored energy of a body by virtue of its position or configuration. For a conservative force F(x), V(x) is defined such that F(x) = -dV/dx.
e.g. Gravitational potential energy of a ball of mass m at height h is V(h) = mgh.
Conservative Force
A force is conservative if the work done by it depends only on the initial and final positions, not on the path taken, and the work done in a closed path is zero.
e.g. Gravitational force and spring force are conservative; friction is non-conservative.
Conservation of Mechanical Energy
The total mechanical energy (sum of kinetic and potential energies) of a system is conserved if the forces doing work on it are conservative. Δ(K + V) = 0.
e.g. A ball dropped from height H: at top, E = mgH; at ground, E = (1/2)mv_f^2; both equal.
Spring Force (Hooke's Law)
The spring force F_s is proportional to the displacement x from equilibrium: F_s = -k x, where k is the spring constant (N/m).
e.g. A car of mass 1000 kg moving at 5 m/s compresses a spring of k = 5.25 × 10^3 N/m by 2.00 m.
Power
Power is the time rate at which work is done or energy is transferred. Average power P_av = W/t, instantaneous power P = dW/dt = F.v.
e.g. An elevator carrying 1800 kg at 2 m/s with friction 4000 N requires power P = 22000 × 2 = 44000 W = 59 hp.
Elastic Collision
A collision in which both total linear momentum and total kinetic energy of the system are conserved.
e.g. Two equal masses colliding elastically: the first stops, the second moves with the initial speed of the first.
Inelastic Collision
A collision in which total linear momentum is conserved but total kinetic energy is not conserved. A completely inelastic collision is one where the two bodies move together after collision.
e.g. A neutron slowing down in a nuclear reactor via inelastic collisions with nuclei.
Formulas & Equations (15)
A.B = A B cos θ
Reads as: A dot B equals A B cos theta
Variables: A, B: magnitudes of vectors; θ: angle between them (degrees or radians)
Use when: Calculating work done by a constant force, or finding the angle between two vectors.
⚠ Using sin θ instead of cos θ, or forgetting that the result is a scalar.
A.B = A_x B_x + A_y B_y + A_z B_z
Reads as: A dot B equals A x B x plus A y B y plus A z B z
Variables: A_x, A_y, A_z: components of vector A; B_x, B_y, B_z: components of vector B
Use when: Computing dot product when vectors are given in component form.
⚠ Adding cross terms like A_x B_y.
W = F.d = F d cos θ
Reads as: Work equals F dot d equals F d cos theta
Variables: F: magnitude of force (N); d: magnitude of displacement (m); θ: angle between force and displacement
Use when: Calculating work done by a constant force.
⚠ Forgetting that displacement must be in the direction of the force component; using the wrong angle.
K = (1/2) m v^2
Reads as: Kinetic energy equals one-half m v squared
Variables: m: mass (kg); v: speed (m/s)
Use when: Calculating kinetic energy of a moving object.
⚠ Using velocity squared but forgetting the 1/2 factor, or using mass in grams.
K_f - K_i = W
Reads as: Final kinetic energy minus initial kinetic energy equals work
Variables: K_f: final kinetic energy (J); K_i: initial kinetic energy (J); W: net work done (J)
Use when: Applying the work-energy theorem to find speed, work, or change in kinetic energy.
⚠ Using work done by a single force instead of net work.
W = ∫_{x_i}^{x_f} F(x) dx
Reads as: Work equals the integral from x i to x f of F of x dx
Variables: F(x): force as a function of position (N); x_i, x_f: initial and final positions (m)
Use when: Calculating work done by a variable force in one dimension.
⚠ Forgetting to integrate or using the wrong limits.
F(x) = -dV/dx
Reads as: F of x equals negative d V by d x
Variables: F(x): conservative force (N); V(x): potential energy function (J)
Use when: Deriving force from potential energy or vice versa.
⚠ Missing the negative sign.
ΔV = -F(x) Δx
Reads as: Change in potential energy equals negative F of x times delta x
Variables: ΔV: change in potential energy (J); F(x): conservative force (N); Δx: displacement (m)
Use when: Relating change in potential energy to work done by a conservative force.
⚠ Forgetting the negative sign.
V(h) = mgh
Reads as: Potential energy at height h equals m g h
Variables: m: mass (kg); g: acceleration due to gravity (≈10 m/s²); h: height above reference (m)
Use when: Calculating gravitational potential energy near Earth's surface.
⚠ Using g = 9.8 when 10 is given, or forgetting that h is measured from a chosen zero.
F_s = -k x
Reads as: Spring force equals negative k x
Variables: k: spring constant (N/m); x: displacement from equilibrium (m)
Use when: Calculating spring force for a given extension or compression.
⚠ Forgetting the negative sign indicating opposite direction.
V(x) = (1/2) k x^2
Reads as: Potential energy of a spring equals one-half k x squared
Variables: k: spring constant (N/m); x: displacement from equilibrium (m)
Use when: Calculating elastic potential energy stored in a spring.
⚠ Forgetting the 1/2 factor.
P = dW/dt = F.v
Reads as: Power equals dW by dt equals F dot v
Variables: P: power (W); W: work (J); t: time (s); F: force (N); v: velocity (m/s)
Use when: Calculating instantaneous power.
⚠ Using average velocity instead of instantaneous velocity.
m_1 v_{1i} = (m_1 + m_2) v_f
Reads as: m1 v1i equals (m1 plus m2) vf
Variables: m_1, m_2: masses (kg); v_{1i}: initial velocity of m1 (m/s); v_f: final common velocity (m/s)
Use when: Solving completely inelastic collisions in one dimension.
⚠ Forgetting that both masses move together after collision.
v_{1f} = (m_1 - m_2)/(m_1 + m_2) v_{1i}
Reads as: v1f equals (m1 minus m2) over (m1 plus m2) times v1i
Variables: v_{1f}: final velocity of m1 (m/s); m_1, m_2: masses (kg); v_{1i}: initial velocity of m1 (m/s)
Use when: Finding final velocity of mass 1 in a one-dimensional elastic collision (m2 initially at rest).
⚠ Swapping masses in the numerator.
v_{2f} = (2 m_1)/(m_1 + m_2) v_{1i}
Reads as: v2f equals (2 m1) over (m1 plus m2) times v1i
Variables: v_{2f}: final velocity of m2 (m/s); m_1, m_2: masses (kg); v_{1i}: initial velocity of m1 (m/s)
Use when: Finding final velocity of mass 2 in a one-dimensional elastic collision (m2 initially at rest).
⚠ Forgetting the factor of 2.
How It Works
Calculating Work Done by a Constant Force
  1. Identify the force vector F and displacement vector d.
  2. Find the angle θ between F and d.
  3. Compute W = F d cos θ, or use dot product W = F.d.
  4. If θ is between 0° and 90°, work is positive; if between 90° and 180°, work is negative; if 90°, work is zero.
💡 Work is zero if displacement is zero, force is zero, or force is perpendicular to displacement.
Applying the Work-Energy Theorem
  1. Identify all forces acting on the object.
  2. Calculate the net work done by all forces (sum of work by each force).
  3. Equate net work to the change in kinetic energy: W_net = ΔK = K_f - K_i.
  4. Solve for the unknown (e.g., final speed, work done by a specific force).
💡 The work-energy theorem applies to the net force, not individual forces.
Calculating Work Done by a Variable Force (1D)
  1. Express the force as a function of position, F(x).
  2. Set up the definite integral W = ∫_{x_i}^{x_f} F(x) dx.
  3. Evaluate the integral (geometrically as area under F-x curve or analytically).
  4. If the force is conservative, the work done equals the negative change in potential energy.
💡 The area under the F-x graph gives the work done; areas below the x-axis are negative.
Conservation of Mechanical Energy
  1. Verify that only conservative forces do work (or non-conservative forces do zero work).
  2. Choose a reference point for zero potential energy.
  3. Write the total mechanical energy at initial position: E_i = K_i + U_i.
  4. Write the total mechanical energy at final position: E_f = K_f + U_f.
  5. Set E_i = E_f and solve for the unknown.
💡 If non-conservative forces (like friction) do work, mechanical energy is not conserved; use the work-energy theorem instead.
Solving Elastic Collisions in One Dimension
  1. Write the conservation of momentum equation: m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}.
  2. Write the conservation of kinetic energy equation: (1/2)m_1 v_{1i}^2 + (1/2)m_2 v_{2i}^2 = (1/2)m_1 v_{1f}^2 + (1/2)m_2 v_{2f}^2.
  3. If one body is initially at rest (v_{2i}=0), use the derived formulas: v_{1f} = (m_1 - m_2)/(m_1 + m_2) v_{1i}, v_{2f} = (2 m_1)/(m_1 + m_2) v_{1i}.
  4. Check special cases: equal masses (m_1 = m_2) → v_{1f}=0, v_{2f}=v_{1i}; heavy target (m_2 >> m_1) → v_{1f} ≈ -v_{1i}, v_{2f} ≈ 0.
💡 In elastic collisions, the relative speed of approach equals the relative speed of separation.
Solving Completely Inelastic Collisions in One Dimension
  1. Write conservation of momentum: m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f.
  2. Solve for the common final velocity: v_f = (m_1 v_{1i} + m_2 v_{2i})/(m_1 + m_2).
  3. Calculate the loss in kinetic energy: ΔK = (1/2) m_1 v_{1i}^2 + (1/2) m_2 v_{2i}^2 - (1/2)(m_1+m_2) v_f^2.
💡 Kinetic energy is not conserved; maximum possible kinetic energy is lost in a completely inelastic collision.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Work with Everyday Effort
✗ A weightlifter holding a 150 kg mass steadily on his shoulder for 30 s does work on the load.
→ In physics, work requires displacement. Since the load is stationary (displacement = 0), no work is done on the load, even though the weightlifter expends energy internally.
✓ No work is done on the load because displacement is zero. The weightlifter's muscles do internal work, but not on the load.
⚠ Sign of Work Done by Friction
✗ Work done by friction is always negative.
→ Friction always opposes relative motion, so the angle between friction and displacement is 180°, giving cos 180° = -1. However, if the object is moving in the same direction as friction (e.g., a block on an accelerating truck), friction can do positive work.
✓ Work done by kinetic friction is usually negative because it opposes motion, but it can be positive if it is the force causing motion (e.g., friction between tyres and road).
⚠ Using Work-Energy Theorem with Non-Conservative Forces
✗ Mechanical energy is conserved even when friction is present.
→ Friction is a non-conservative force; it dissipates mechanical energy into heat. The work-energy theorem still holds (ΔK = W_net), but the conservation of mechanical energy (ΔK + ΔU = 0) does not.
✓ When non-conservative forces like friction act, mechanical energy is not conserved. Use the work-energy theorem: ΔK = W_conservative + W_non-conservative.
⚠ Forgetting the Negative Sign in Spring Force
✗ Spring force F_s = k x.
→ Hooke's law states F_s = -k x. The negative sign indicates that the spring force always opposes the displacement (restoring force). Omitting the sign gives the wrong direction.
✓ F_s = -k x, where x is displacement from equilibrium. The negative sign shows the force is opposite to displacement.
⚠ Confusing Elastic and Inelastic Collisions
✗ In an inelastic collision, momentum is not conserved.
→ Momentum is conserved in all collisions (elastic and inelastic) because no external force acts during the collision. Only kinetic energy is not conserved in inelastic collisions.
✓ In all collisions, total linear momentum is conserved. In elastic collisions, kinetic energy is also conserved; in inelastic collisions, kinetic energy is not conserved.
⚠ Misapplying Power Formula
✗ Power = F × v, using average velocity.
→ The formula P = F.v gives instantaneous power. If force is constant, average power = F × v_avg, but if force varies, you must use instantaneous values or integrate.
✓ Instantaneous power P = F.v. For constant force and constant velocity, average power = F × v. For variable force, use P = dW/dt.
Compare & Contrast
Work done in a closed path
Conservative Force
Non-Conservative Force
Work done by a conservative force in a closed path is zero; work done by a non-conservative force (e.g., friction) in a closed path is not zero.
Conservation of kinetic energy
Elastic Collision
Inelastic Collision
In an elastic collision, total kinetic energy is conserved; in an inelastic collision, total kinetic energy is not conserved (some is converted to heat, sound, etc.).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11 marks.
Most asked: Work done by constant and variable forces · Work-energy theorem · Conservation of mechanical energy (free fall, pendulum, spring) · Elastic and inelastic collisions in one dimension · Power calculations
Don't skip: Numerical on work-energy theorem with variable force or friction; Derivation or numerical on conservation of mechanical energy (e.g., pendulum, spring-block system); Collision problems (elastic and inelastic) with momentum and energy conservation
One-Line Revision (read aloud before the exam)
  1. Work done by a constant force is W = F d cos θ, where θ is the angle between force and displacement.
  2. Kinetic energy is K = ½ mv², a scalar measure of motion-related energy.
  3. The work-energy theorem states that the net work done on a particle equals its change in kinetic energy.
  4. Work done by a variable force is the area under the force-displacement graph.
  5. Potential energy is stored energy due to position or configuration, defined only for conservative forces.
  6. Gravitational potential energy near Earth is U = mgh, with h measured from a chosen zero.
  7. Spring potential energy is U = ½ kx², where x is displacement from equilibrium.
  8. Total mechanical energy (K + U) is conserved if only conservative forces do work.
  9. Power is the rate of doing work: P = dW/dt = F.v, measured in watts (W).
  10. In all collisions, total linear momentum is conserved.
  11. In elastic collisions, total kinetic energy is conserved; in inelastic collisions, it is not.
  12. A completely inelastic collision is one where the colliding bodies stick together after impact.
  13. The scalar product of two vectors is commutative: A.B = B.A.
  14. For unit vectors: î.î = ĵ.ĵ = k̂.k̂ = 1, and î.ĵ = ĵ.k̂ = k̂.î = 0.
  15. The work done by a conservative force depends only on initial and final positions, not the path.

MCQs — 10 Questions with Answers

Q1. Two vectors A = 3î + 4ĵ and B = 4î + 3ĵ are given. What is their scalar product A·B?

  • A. 24 ✓
  • B. 25
  • C. 7
  • D. 1

Answer: A — A·B = (3)(4) + (4)(3) + (0)(0) = 12 + 12 = 24.

Q2. A force of 10 N acts on an object at an angle of 60° to its displacement of 5 m. Calculate the work done.

  • A. 25 J ✓
  • B. 50 J
  • C. 75 J
  • D. 100 J

Answer: A — W = Fd cos θ = 10 × 5 × cos 60° = 10 × 5 × 0.5 = 25 J.

Q3. A 2 kg object accelerates from 3 m/s to 5 m/s. What is the change in its kinetic energy?

  • A. 8 J
  • B. 16 J ✓
  • C. 20 J
  • D. 4 J

Answer: B — ΔKE = ½ × m × (vf² – vi²) = ½ × 2 × (5² – 3²) = 1 × (25 – 9) = 16 J.

Q4. Which of the following statements about work done by friction is correct?

  • A. Work by friction is always positive because friction is a force
  • B. Work by friction is always negative because it opposes motion ✓
  • C. Work by friction equals zero if the object moves in a circle
  • D. Work by friction depends only on the friction force, not on displacement

Answer: B — Friction opposes motion, so θ = 180° and cos 180° = −1, making W = −Fd always negative.

Q5. A raindrop of mass 1 g falls from height 1 km and hits ground at 50 m/s. If gravitational work is 10 J, what is work done by air resistance? (Take g = 10 m/s²)

  • A. −8.75 J ✓
  • B. 1.25 J
  • C. 11.25 J
  • D. 8.75 J

Answer: A — ΔKE = ½(0.001)(50²) = 1.25 J; from Wnet = ΔKE: Wair = ΔKE − Wgrav = 1.25 − 10 = −8.75 J.

Q6. A force F = 3î + 4ĵ − 5k̂ (in N) acts on an object with displacement d = 5î + 4ĵ + 3k̂ (in m). Find the work done.

  • A. 16 J ✓
  • B. 40 J
  • C. 30 J
  • D. 12 J

Answer: A — W = F·d = (3)(5) + (4)(4) + (−5)(3) = 15 + 16 − 15 = 16 J.

Q7. Which statement is NOT correct regarding scalar product of perpendicular vectors?

  • A. Scalar product equals zero
  • B. cos 90° term makes product zero
  • C. Dot product is undefined for perpendicular vectors ✓
  • D. Example: î·ĵ = 0

Answer: C — Dot product is always defined; for perpendicular vectors it simply equals zero because cos 90° = 0.

Q8. A cyclist applies force to move 10 m in 5 seconds doing 200 J of work. Which scenario requires MORE power? (Both cases: same 200 J work done)

  • A. Moving 10 m in 2 seconds (P = 100 W) ✓
  • B. Moving 10 m in 5 seconds (P = 40 W)
  • C. Moving 10 m in 10 seconds (P = 20 W)
  • D. Moving 20 m in 5 seconds (P = 200 W)

Answer: A — Power P = W/t; same work done in less time means higher power; P = 200/2 = 100 W is maximum.

Q9. An object of mass m moves under net force with kinetic energy increasing from 10 J to 30 J. Which statement about work-energy theorem is correct?

  • A. Net work done equals 20 J and object must be speeding up ✓
  • B. Net work done equals −20 J and object slows down
  • C. Work depends on the path taken by the object
  • D. Work done depends only on initial and final positions, not on applied forces

Answer: A — Work-Energy Theorem: Wnet = ΔKE = 30 − 10 = 20 J; positive work increases KE, so object speeds up.

Q10. A 4 kg block slides down a rough incline losing 5 J to friction while gravitational force does 100 J of work. Using work-energy theorem, find change in kinetic energy. (Assume no other forces)

  • A. 95 J ✓
  • B. 105 J
  • C. −5 J
  • D. 100 J

Answer: A — Wnet = Wgrav + Wfriction = 100 + (−5) = 95 J; by work-energy theorem, ΔKE = Wnet = 95 J.

Flashcards

Define scalar product (dot product) of two vectors A and B.

A.B = AB cos θ, where θ is the angle between vectors; result is a scalar quantity with no direction.

What is work done by a constant force?

Work W = F.d = Fd cos θ, where θ is the angle between force and displacement; SI unit is joule (J).

Write the kinetic energy formula and state its key property.

KE = ½mv²; it is always non-negative and depends only on speed, not direction of motion.

State the Work-Energy Theorem in one sentence.

The net work done on a particle equals its change in kinetic energy: W(net) = ΔKE = KE(final) − KE(initial).

Why is work done by friction on a sliding object negative?

Friction force opposes motion, so the angle θ = 180°, making cos 180° = −1 and W = −Fd.

What is the geometric meaning of B cos θ in the scalar product A.B?

B cos θ is the projection (component) of vector B onto the direction of vector A.

When is work done by a force zero?

Work is zero when force is perpendicular to displacement (θ = 90°), because cos 90° = 0.

How do you calculate work done by a variable force?

Work = integral of F.d over the displacement path; for constant force in one direction, W = Fd cos θ.

What is the relationship between power, work, and time?

Power P = W/t; it measures the rate at which work is done; SI unit is watt (W or J/s).

Explain why dot product of perpendicular unit vectors is zero.

î·ĵ = |î||ĵ| cos 90° = 1 × 1 × 0 = 0, because perpendicular vectors have θ = 90°.

Important Board Questions

Define work done by a constant force. If a force of 5 N acts at 30° to a displacement of 4 m, calculate the work done. [2 marks]

Use W = Fd cos θ; substitute F = 5 N, d = 4 m, θ = 30°; cos 30° = √3/2 ≈ 0.866.

A 1500 kg car accelerates from rest to 20 m/s in 10 seconds on a straight road. (a) Calculate the change in kinetic energy. (b) Using the work-energy theorem, find the net work done by the engine. [5 marks]

Part (a): Use ΔKE = ½m(vf² − vi²) with vi = 0. Part (b): By work-energy theorem, Wnet = ΔKE calculated in part (a); show all substitutions and final numerical answer.

Derive the work-energy theorem starting from Newton's second law (F = ma) and the kinematic equation (v² − u² = 2as). Show that the net work done on a particle equals its change in kinetic energy. [6 marks]

Start with v² − u² = 2as; multiply both sides by m/2; use F = ma to replace ma; rearrange to show ½mv² − ½mu² = Fs, which is Wnet = ΔKE; explain the physical meaning of each term in the final equation.

Assertion-Reason Questions for Work, Energy and Power

Q1. Assertion (A):

The scalar product A·B = AB cos θ is commutative, meaning A·B = B·A.

Reason (R):

The magnitude of A and B remain the same regardless of the order, and cos θ is the same angle between both vectors in either order.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. The scalar product is indeed commutative because multiplication of scalars (the magnitudes and cosine) is commutative in mathematics. The reason correctly explains why—the angle between A and B is the same whether calculated as A to B or B to A, and magnitudes are independent of order.

Q2. Assertion (A):

The work done by a force perpendicular to displacement is always zero.

Reason (R):

When θ = 90°, cos 90° = 0, so W = Fd cos 90° = 0.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. The mathematical derivation from the work formula directly proves that perpendicular forces do no work. This is a fundamental principle in physics.

Q3. Assertion (A):

The change in kinetic energy of an object equals the total work done on it by all forces, as stated by the work-energy theorem.

Reason (R):

Kinetic energy is a vector quantity that depends on the direction of motion.

Answer: C — A is true but R is false — The assertion is true—this is the correct statement of the work-energy theorem. However, the reason is false. Kinetic energy is a scalar quantity (K = ½mv²), not a vector. It depends only on the magnitude of velocity, not its direction. The reason provides an incorrect explanation for a correct assertion.

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Case-Based Questions (CBQ) for Work, Energy and Power

Case Study:

A delivery truck of mass 5000 kg accelerates uniformly from rest on a straight, level road. After traveling 50 m, it reaches a speed of 10 m/s. During this acceleration, the engine provides a forward force, and resistance forces (air drag and rolling friction) total 2000 N. An engineer wants to calculate the net work done on the truck and the power output.

Q1. What is the change in kinetic energy of the truck as it accelerates from 0 to 10 m/s?

  • A. 50,000 J
  • B. 100,000 J
  • C. 250,000 J ✓
  • D. 500,000 J

Answer: C

Q2. If the truck travels 50 m during this acceleration, what is the net force acting on it?

  • A. 5000 N
  • B. 2500 N ✓
  • C. 10,000 N
  • D. 1000 N

Answer: B

Q3. Using the work-energy theorem, what is the total work done by all forces on the truck?

  • A. 100,000 J
  • B. 125,000 J
  • C. 250,000 J ✓
  • D. 500,000 J

Answer: C

Q4. If the truck takes 10 seconds to travel the 50 m, what is the average power output of the engine (considering the engine must overcome resistance)?

  • A. 25,000 W
  • B. 27,500 W ✓
  • C. 35,000 W
  • D. 50,000 W

Answer: B

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