Work, Energy, and Power — Hard Practice Quiz

A Physics cheat sheet for Work, Energy, and Power — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Formulas & key concepts

Work Done by a Constant Force: The work \(W\) done by a force \(F\) over a displacement \(d\) at an angle \(\theta\).

$$W = Fd\cos(\theta)$$

Work-Energy Theorem: The net work \(W_{net}\) done on an object equals its change in kinetic energy \(\Delta KE\).

$$W_{net} = \Delta KE$$

Kinetic Energy: The energy \(KE\) of an object due to its motion, where \(m\) is mass and \(v\) is velocity.

$$KE = \frac{1}{2}mv^2$$

Gravitational Potential Energy: Energy \(PE_g\) stored based on vertical position \(h\) in a gravitational field.

$$PE_g = mgh$$

Where: \(g\) = acceleration due to gravity (approx. \(9.8 m/s^2\) on Earth)

Elastic Potential Energy: Energy \(PE_s\) stored in a spring stretched/compressed by distance \(x\).

$$PE_s = \frac{1}{2}kx^2$$

Where: \(k\) = spring constant

Total Mechanical Energy: Sum of kinetic and potential energies.

$$E_{total} = KE + PE_g + PE_s$$

Power: The rate at which work \(W\) is done or energy is transferred over time \(t\).

$$P = \frac{W}{t}$$

Work Done by Gravity: The work done by the gravitational force as an object moves from initial height \(h_0\) to final height \(h_f\).

$$W_{gravity} = mg(h_0 - h_f)$$

Work by Non-conservative Forces: The net work \(W_{nc}\) done by non-conservative forces equals the change in total mechanical energy.

$$W_{nc} = E_f - E_0$$

Where: \(E\) = total mechanical energy (KE + PE)

Power (Energy Change): Power is also defined as the rate at which energy changes.

$$P = \frac{\Delta E}{t}$$

Work Done by a Variable Force: The work done is equal to the area under the graph of the force component \(F\cos\theta\) versus displacement \(s\).

$$W = \text{Area under } F\cos\theta \text{ vs } s$$

Instantaneous Power: The product of force \(F\) and velocity \(v\) in the direction of the force.

$$P = Fv\cos(\theta)$$

Practice quiz

  1. A $2.0 \text{ kg}$ block is initially at rest on a frictionless horizontal surface. A constant horizontal force of $10 \text{ N}$ is applied to the block for $4.0 \text{ s}$. What is the average power delivered by the force during this time?

    • $50 \text{ W}$
    • $100 \text{ W}$
    • $200 \text{ W}$
    • $400 \text{ W}$

    Answer: $100 \text{ W}$

  2. A $0.5 \text{ kg}$ ball is dropped from a height of $10 \text{ m}$. It strikes the ground and rebounds to a height of $6 \text{ m}$. If the collision with the ground lasts for $0.02 \text{ s}$, what is the average power dissipated by non-conservative forces during the collision? (Assume $g = 9.8 \text{ m/s}^2$)

    • $490 \text{ W}$
    • $980 \text{ W}$
    • $1470 \text{ W}$
    • $1960 \text{ W}$

    Answer: $980 \text{ W}$

  3. A block of mass $m$ is released from rest at a height $H$ on a frictionless incline. It slides down and compresses a spring with spring constant $k$ by a maximum distance $x$. Assuming the reference height for gravitational potential energy is the lowest point of spring compression, which equation correctly relates $x$ to $m$, $g$, $H$, and $k$?

    • $\frac{1}{2}kx^2 = mgH$
    • $\frac{1}{2}kx^2 = mg(H+x)$
    • $mgH = \frac{1}{2}kx^2 + mgx$
    • $mg(H-x) = \frac{1}{2}kx^2$

    Answer: $\frac{1}{2}kx^2 = mg(H+x)$

  4. A pump lifts $500 \text{ kg}$ of water per minute from a well $20 \text{ m}$ deep and ejects it with a speed of $10 \text{ m/s}$. What is the minimum power required by the pump? (Assume $g = 9.8 \text{ m/s}^2$)

    • $1633 \text{ W}$
    • $2050 \text{ W}$
    • $2450 \text{ W}$
    • $2950 \text{ W}$

    Answer: $2050 \text{ W}$

  5. An object of mass $m$ is initially at rest. A constant force $F$ acts on it, causing it to accelerate over a distance $d$. If the mass of the object is doubled, but the force and distance remain the same, how do the final kinetic energy and the average power delivered by the force change?

    • Kinetic energy doubles, average power remains the same.
    • Kinetic energy remains the same, average power decreases by a factor of $\sqrt{2}$.
    • Kinetic energy remains the same, average power increases by a factor of $\sqrt{2}$.
    • Kinetic energy doubles, average power decreases by a factor of $\sqrt{2}$.

    Answer: Kinetic energy remains the same, average power decreases by a factor of $\sqrt{2}$.

  6. A spring with spring constant $k$ is compressed by a distance $x$. It then launches a ball of mass $m$ vertically upwards. If the ball reaches a maximum height $h$ above the uncompressed spring position, which of the following expressions correctly relates $k$, $x$, $m$, $g$, and $h$?

    • $\frac{1}{2}kx^2 = mgh$
    • $kx = mgh$
    • $\frac{1}{2}kx^2 = \frac{1}{2}mv^2 + mgh$
    • $kx^2 = mgh$

    Answer: $\frac{1}{2}kx^2 = mgh$

  7. A $1500 \text{ kg}$ car accelerates from rest to $20 \text{ m/s}$ in $5.0 \text{ s}$. During this time, a constant friction force of $300 \text{ N}$ acts on the car. What is the average power delivered by the engine?

    • $60,000 \text{ W}$
    • $63,000 \text{ W}$
    • $66,000 \text{ W}$
    • $75,000 \text{ W}$

    Answer: $63,000 \text{ W}$

  8. A block of mass $m$ is pulled up a rough incline (angle $\alpha$) by a constant force $F$ parallel to the incline. The block moves a distance $d$ up the incline at a constant velocity. If the coefficient of kinetic friction is $\mu_k$, what is the work done by the applied force $F$?

    • $mgd\sin(\alpha)$
    • $\mu_k mgd\cos(\alpha)$
    • $(\mu_k mg\cos(\alpha) + mg\sin(\alpha)) d$
    • $(F - \mu_k mg\cos(\alpha) - mg\sin(\alpha)) d$

    Answer: $(\mu_k mg\cos(\alpha) + mg\sin(\alpha)) d$

  9. An object is launched vertically upwards with an initial kinetic energy $KE_0$. Ignoring air resistance, at what height will its kinetic energy be exactly one-third of its initial kinetic energy? Express your answer in terms of $KE_0$, $m$, and $g$.

    • $h = \frac{KE_0}{3mg}$
    • $h = \frac{2KE_0}{3mg}$
    • $h = \frac{KE_0}{2mg}$
    • $h = \frac{3KE_0}{2mg}$

    Answer: $h = \frac{2KE_0}{3mg}$

  10. A variable force acts on a $2.0 \text{ kg}$ object, causing it to move along the x-axis. The force is given by $F(x) = (3x^2 - 2x) \text{ N}$, where $x$ is in meters. If the object starts from rest at $x=0$, what is its speed when it reaches $x=2.0 \text{ m}$?

    • $1.0 \text{ m/s}$
    • $2.0 \text{ m/s}$
    • $3.0 \text{ m/s}$
    • $4.0 \text{ m/s}$

    Answer: $2.0 \text{ m/s}$

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