Energy Practice Problems

Problem 1: Work-Energy Theorem

1. A 5 kg box is pushed up a 4 m incline that makes a 30° angle with the horizontal. A constant force of 40 N is applied parallel to the incline. The coefficient of kinetic friction between the box and the incline is 0.2.
   a) Calculate the work done by the applied force.
   b) Determine the work done by friction.
   c) Using the work-energy theorem, find the final speed of the box if it starts from rest.

Problem 2: Conservation of Mechanical Energy

1. A 0.2 kg ball is dropped from a height of 10 m.
   a) Calculate the potential energy of the ball relative to the ground before it is released.
   b) Determine the speed of the ball just before it hits the ground, assuming no air resistance.
   c) Explain how the conservation of mechanical energy applies to this scenario.

Problem 3: Spring Potential Energy and Conservation of Energy

1. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. The spring is compressed by 0.1 m from its equilibrium position and then released.
   a) Calculate the initial potential energy stored in the spring.
   b) Determine the speed of the mass when it passes through the equilibrium position.
   c) Discuss how the energy transformation occurs as the mass moves from the compressed position to the equilibrium position.

Problem 4: Gravitational Potential Energy in Space

1. A satellite with a mass of 500 kg orbits Earth at a height of 500 km above the surface. The gravitational constant is G=6.674×10−11 N⋅m2/kg2G = 6.674 \times 10^{-11} \, \text{N}\cdot \text{m}^2/\text{kg}^2G=6.674×10−11N⋅m2/kg2, and Earth's mass is 5.98×1024 kg5.98 \times 10^{24} \, \text{kg}5.98×1024kg. Earth's radius is approximately 6.37×106 m6.37 \times 10^6 \, \text{m}6.37×106m.
   a) Calculate the gravitational potential energy of the satellite in orbit relative to Earth.
   b) Determine the satellite’s total mechanical energy in orbit.
   c) Explain why the satellite does not need additional energy to maintain its orbit.

Problem 5: Power and Work

1. A 1,200 kg car accelerates from 0 to 25 m/s in 10 seconds.
   a) Calculate the work done on the car to reach this speed.
   b) Determine the average power output of the engine during this acceleration.
   c) If the engine were only 60% efficient, calculate the actual power output required from the fuel.

Problem 6: Elastic and Inelastic Collisions

1. A 2 kg ball moving at 4 m/s collides elastically with a stationary 1 kg ball.
   a) Calculate the velocities of both balls after the collision.
   b) Determine the kinetic energy of each ball before and after the collision.
   c) Discuss how conservation of kinetic energy applies to elastic collisions.

Problem 7: Work Done by a Variable Force

1. A force F(x)=3x2F(x) = 3x^2F(x)=3x2 N acts on a 1 kg particle as it moves along the x-axis from x=0x = 0x=0 to x=4x = 4x=4 m.
   a) Calculate the work done by this force over the distance traveled.
   b) If the particle starts from rest, find its final speed at x=4x = 4x=4 m.
   c) Explain the relationship between work done and kinetic energy in this case.

Problem 8: Energy in a Pendulum System

1. A pendulum with a 0.5 kg bob swings from a height of 0.4 m above its lowest point.
   a) Calculate the potential energy of the bob at its highest point relative to the lowest point.
   b) Determine the bob’s speed as it passes through the lowest point.
   c) Describe the energy transformations that occur during the pendulum's swing.

Problem 9: Energy Conservation with Friction

1. A 3 kg block slides down a 5 m high frictionless incline, then moves across a horizontal surface with a coefficient of kinetic friction of 0.3 until it comes to a stop.
   a) Calculate the block’s speed at the bottom of the incline.
   b) Determine the distance it travels on the horizontal surface before coming to rest.
   c) Discuss how energy conservation applies to this scenario, considering the work done by friction.

Problem 10: Potential Energy and Escape Velocity

1. A projectile of mass 0.5 kg is launched straight up from Earth’s surface. Ignoring air resistance, calculate the following:
   a) The gravitational potential energy of the projectile at a height of 1,000 km above Earth’s surface.
   b) The minimum initial speed required for the projectile to reach this height.
   c) Discuss how gravitational potential energy and kinetic energy change as the projectile ascends.

Answer Key for Energy Practice Problems

Problem 1: Work-Energy Theorem

1. a) Work done by the applied force:
   W=F⋅d=40×4=160 JW = F \cdot d = 40 \times 4 = 160 \, \text{J}W=F⋅d=40×4=160J

   b) Work done by friction:
   Frictional force, fk=μk⋅m⋅g⋅cos⁡(30∘)≈0.2×5×9.8×cos⁡(30∘)=8.49 Nf_k = \mu_k \cdot m \cdot g \cdot \cos(30^\circ) \approx 0.2 \times 5 \times 9.8 \times \cos(30^\circ) = 8.49 \, \text{N}fk​=μk​⋅m⋅g⋅cos(30∘)≈0.2×5×9.8×cos(30∘)=8.49N
   Work by friction, Wf=fk⋅d=8.49×4=33.96 JW_f = f_k \cdot d = 8.49 \times 4 = 33.96 \, \text{J}Wf​=fk​⋅d=8.49×4=33.96J

   c) Final speed of the box:
   Net work done Wnet=160−33.96=126.04 JW_{\text{net}} = 160 - 33.96 = 126.04 \, \text{J}Wnet​=160−33.96=126.04J
   Using Wnet=ΔKE=12mv2W_{\text{net}} = \Delta KE = \frac{1}{2} m v^2Wnet​=ΔKE=21​mv2, solve for vvv:
   v=2⋅126.045≈7.1 m/sv = \sqrt{\frac{2 \cdot 126.04}{5}} \approx 7.1 \, \text{m/s}v=52⋅126.04​​≈7.1m/s

Problem 2: Conservation of Mechanical Energy

1. a) Potential energy before release:
   U=mgh=0.2×9.8×10=19.6 JU = mgh = 0.2 \times 9.8 \times 10 = 19.6 \, \text{J}U=mgh=0.2×9.8×10=19.6J

   b) Speed just before hitting the ground:
   KE=U⇒12mv2=19.6KE = U \Rightarrow \frac{1}{2} mv^2 = 19.6KE=U⇒21​mv2=19.6
   v=2×19.60.2≈14 m/sv = \sqrt{\frac{2 \times 19.6}{0.2}} \approx 14 \, \text{m/s}v=0.22×19.6​​≈14m/s

   c) Conservation of mechanical energy:
   The initial potential energy is converted into kinetic energy as the ball falls, with total mechanical energy remaining constant.

Problem 3: Spring Potential Energy and Conservation of Energy

1. a) Initial potential energy in the spring:
   U=12kx2=12×200×(0.1)2=1 JU = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 1 \, \text{J}U=21​kx2=21​×200×(0.1)2=1J

   b) Speed at equilibrium position:
   KE=U⇒12mv2=1KE = U \Rightarrow \frac{1}{2} mv^2 = 1KE=U⇒21​mv2=1
   v=2×11=1.41 m/sv = \sqrt{\frac{2 \times 1}{1}} = 1.41 \, \text{m/s}v=12×1​​=1.41m/s

   c) Energy transformation:
   Potential energy stored in the compressed spring is fully converted to kinetic energy as the mass moves through the equilibrium position.

Problem 4: Gravitational Potential Energy in Space

1. a) Gravitational potential energy:
   U=−GMmrU = -\frac{GMm}{r}U=−rGMm​, where r=R+h=6.37×106+5×105≈6.87×106 mr = R + h = 6.37 \times 10^6 + 5 \times 10^5 \approx 6.87 \times 10^6 \, \text{m}r=R+h=6.37×106+5×105≈6.87×106m
   U=−6.674×10−11×5.98×1024×5006.87×106≈−2.9×1010 JU = -\frac{6.674 \times 10^{-11} \times 5.98 \times 10^{24} \times 500}{6.87 \times 10^6} \approx -2.9 \times 10^{10} \, \text{J}U=−6.87×1066.674×10−11×5.98×1024×500​≈−2.9×1010J

   b) Total mechanical energy:
   E=12U≈−1.45×1010 JE = \frac{1}{2} U \approx -1.45 \times 10^{10} \, \text{J}E=21​U≈−1.45×1010J

   c) Explanation:
   The satellite’s orbit requires no additional energy since gravitational force provides the centripetal force needed for stable circular motion.

Problem 5: Power and Work

1. a) Work done:
   W=ΔKE=12mv2=12×1200×252=3.75×105 JW = \Delta KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times 25^2 = 3.75 \times 10^5 \, \text{J}W=ΔKE=21​mv2=21​×1200×252=3.75×105J

   b) Average power output:
   P=Wt=3.75×10510=3.75×104 WP = \frac{W}{t} = \frac{3.75 \times 10^5}{10} = 3.75 \times 10^4 \, \text{W}P=tW​=103.75×105​=3.75×104W

   c) Actual power output at 60% efficiency:
   Pactual=3.75×1040.6≈6.25×104 WP_{\text{actual}} = \frac{3.75 \times 10^4}{0.6} \approx 6.25 \times 10^4 \, \text{W}Pactual​=0.63.75×104​≈6.25×104W

Problem 6: Elastic and Inelastic Collisions

1. a) Velocities after collision:
   Using conservation of momentum and kinetic energy, v1f=1.33 m/sv_{1f} = 1.33 \, \text{m/s}v1f​=1.33m/s (for 2 kg ball), and v2f=5.33 m/sv_{2f} = 5.33 \, \text{m/s}v2f​=5.33m/s (for 1 kg ball).

   b) Kinetic energy:
   Before: KEtotal=12⋅2⋅42=16 JKE_{\text{total}} = \frac{1}{2} \cdot 2 \cdot 4^2 = 16 \, \text{J}KEtotal​=21​⋅2⋅42=16J
   After: KEtotal=12⋅2⋅(1.33)2+12⋅1⋅(5.33)2=16 JKE_{\text{total}} = \frac{1}{2} \cdot 2 \cdot (1.33)^2 + \frac{1}{2} \cdot 1 \cdot (5.33)^2 = 16 \, \text{J}KEtotal​=21​⋅2⋅(1.33)2+21​⋅1⋅(5.33)2=16J

   c) Explanation:
   In an elastic collision, total kinetic energy is conserved.

Problem 7: Work Done by a Variable Force

1. a) Work done:
   W=∫043x2 dx=[x3]04=64 JW = \int_{0}^{4} 3x^2 \, dx = [x^3]_{0}^{4} = 64 \, \text{J}W=∫04​3x2dx=[x3]04​=64J

   b) Final speed at x=4 mx = 4 \, \text{m}x=4m:
   KE=W⇒12mv2=64KE = W \Rightarrow \frac{1}{2} mv^2 = 64KE=W⇒21​mv2=64
   v=2×641=11.3 m/sv = \sqrt{\frac{2 \times 64}{1}} = 11.3 \, \text{m/s}v=12×64​​=11.3m/s

   c) Explanation:
   The work done by the variable force changes the kinetic energy, resulting in acceleration.

Problem 8: Energy in a Pendulum System

1. a) Potential energy at the highest point:
   U=mgh=0.5×9.8×0.4=1.96 JU = mgh = 0.5 \times 9.8 \times 0.4 = 1.96 \, \text{J}U=mgh=0.5×9.8×0.4=1.96J

   b) Speed at the lowest point:
   KE=U⇒12mv2=1.96KE = U \Rightarrow \frac{1}{2} mv^2 = 1.96KE=U⇒21​mv2=1.96
   v=2×1.960.5=2.8 m/sv = \sqrt{\frac{2 \times 1.96}{0.5}} = 2.8 \, \text{m/s}v=0.52×1.96​​=2.8m/s

   c) Energy transformations:
   Potential energy is converted into kinetic energy as the bob swings downward and back to potential energy as it swings upward.

Problem 9: Energy Conservation with Friction

1. a) Speed at the bottom of the incline:
   KE=mgh=3×9.8×5=147 JKE = mgh = 3 \times 9.8 \times 5 = 147 \, \text{J}KE=mgh=3×9.8×5=147J
   v=2×1473=9.8 m/sv = \sqrt{\frac{2 \times 147}{3}} = 9.8 \, \text{m/s}v=32×147​​=9.8m/s

   b) Distance on horizontal surface:
   Frictional force, fk=μk⋅mg=0.3×3×9.8=8.82 Nf_k = \mu_k \cdot mg = 0.3 \times 3 \times 9.8 = 8.82 \, \text{N}fk​=μk​⋅mg=0.3×3×9.8=8.82N
   d=KEfk=1478.82≈16.7 md = \frac{KE}{f_k} = \frac{147}{8.82} \approx 16.7 \, \text{m}d=fk​KE​=8.82147​≈16.7m

   c) Explanation:
   Initial potential energy converts to kinetic energy on the incline and is gradually dissipated by friction on the horizontal surface.

Problem 10: Potential Energy and Escape Velocity

1. a) Gravitational potential energy at 1,000 km:
   U=−GMmrU = -\frac{GMm}{r}U=−rGMm​, where r=R+h=6.37×106+106=7.37×106 mr = R + h = 6.37 \times 10^6 + 10^6 = 7.37 \times 10^6 \, \text{m}r=R+h=6.37×106+106=7.37×106m
   U=−6.674×10−11×5.98×1024×0.57.37×106≈−2.71×107 JU = -\frac{6.674 \times 10^{-11} \times 5.98 \times 10^{24} \times 0.5}{7.37 \times 10^6} \approx -2.71 \times 10^7 \, \text{J}U=−7.37×1066.674×10−11×5.98×1024×0.5​≈−2.71×107J

b) Minimum initial speed:
Using KE=−ΔUKE = -\Delta UKE=−ΔU:
v=2×2.71×1070.5≈10.4 km/sv = \sqrt{\frac{2 \times 2.71 \times 10^7}{0.5}} \approx 10.4 \, \text{km/s}v=0.52×2.71×107​​≈10.4km/s

c) Explanation of energy change:
As the projectile ascends, kinetic energy decreases while gravitational potential energy increases, conserving total mechanical energy.