Questions: Work, Power, and Energy: Fundamental Definitions

Question 1

A 400 N crate slides 5 meters across a frictionless horizontal floor. The normal force from the floor acts upward with magnitude 400 N throughout the motion. How much work does the normal force do?

• A. 2000 J — the normal force equals the weight and acts over the full 5-meter displacement
• B. 0 J — the normal force is perpendicular to the displacement, so the dot product is zero
• C. −2000 J — the normal force opposes the downward weight, doing negative work
• D. 400 J — work equals force magnitude regardless of direction

Answer: (B)

Work is W = ∫ F · dr — the dot product of force and displacement. The normal force acts vertically (upward) while the displacement is horizontal. These are perpendicular, so F · dr = F·dr·cos(90°) = 0. The normal force does zero work, even though it is large and the object is moving. Only force components along the direction of motion contribute to work. This is a critical point: a large force is not sufficient to do work — direction matters.

Question 2

A motor outputs 45 kW of power while driving a conveyor belt. The belt moves at 15 m/s. What force does the motor exert on the belt?

• A. 675,000 N — multiply power (W) by speed (m/s) to get force
• B. 3000 N — use F = P/v = 45,000/15
• C. 30 N — the power in kW divided by the speed gives force in N directly
• D. 45 N — each kW of power produces 1 N of force per m/s

Answer: (B)

Power P = F · v, so F = P/v = 45,000 W / 15 m/s = 3000 N. The common error (option A) multiplies instead of divides — it confuses the relationship. Always convert power to watts before dividing by speed in m/s to obtain force in newtons.

Question 3

A force acting on a moving object can do zero work on that object even if the force is nonzero and the object moves a nonzero distance.

• T. True
• F. False

Answer: True

Work is W = F · d · cos(θ), where θ is the angle between the force and displacement vectors. If θ = 90° (force perpendicular to motion), cos(90°) = 0 and the work is zero regardless of the force magnitude or distance traveled. Common examples include the normal force on a horizontal surface and centripetal force in circular motion — both act perpendicular to velocity and do no work.

Question 4

The work done on an object by a constant force equals the magnitude of that force multiplied by the total distance the object travels.

• T. True
• F. False

Answer: False

Work equals the force component along the displacement, multiplied by the displacement: W = F·d·cos(θ). If the force is at an angle to the motion, only its component along the displacement does work — the perpendicular component contributes nothing. For example, a force of 100 N applied at 60° to the direction of motion over 10 m does only 100 × 10 × cos(60°) = 500 J, not 1000 J. 'Force times distance' is only correct when the force is parallel to the displacement.

Question 5

Why does the work-energy theorem allow you to solve for a particle's final speed using only the net work done and the initial speed, without tracking what forces did at intermediate points along the path?

Model answer: The work-energy theorem states that W_net = ΔKE = ½mv_f² − ½mv_i². Because work is defined as the integral of the net force along the displacement — a scalar quantity that accumulates all force contributions over the entire path — it condenses the entire force history into a single number. The theorem then maps that accumulated work directly to the change in kinetic energy, which depends only on initial and final speeds. The path shape, the variation in force along the way, and intermediate velocities are all already captured in the work integral and do not need to be tracked separately.

This is what makes energy methods more efficient than applying F = ma directly: instead of solving a differential equation for velocity as a function of position, you integrate force over displacement once (or recognize W = Fd for constant forces) and get the speed change immediately.