Energy and Work Calculations

Practice Calculations: Energy

Work Done

The work done (WW) is calculated using the formula:

W=F×dW = F \times d

where:

  • WW is the work done (in joules, J)
  • FF is the force applied (in newtons, N)
  • dd is the distance over which the force is applied (in meters, m)

Example 1: A man pushes a box with a force of 150 N over a distance of 4 m. The work done is:

W=150 N×4 m=600 JW = 150 \text{ N} \times 4 \text{ m} = 600 \text{ J}

Example 2: A student pulls a trolley with a force of 80 N for a distance of 10 m. The work done is:

W=80 N×10 m=800 JW = 80 \text{ N} \times 10 \text{ m} = 800 \text{ J}

Kinetic Energy

Kinetic energy (KEKE) is the energy of motion and is calculated as:

KE=12mv2KE = \frac{1}{2}mv^2

where:

  • KEKE is the kinetic energy (in joules, J)
  • mm is the mass of the object (in kilograms, kg)
  • vv is the velocity of the object (in meters per second, m/s)

Example 3: A car of mass 1,200 kg is moving at a speed of 20 m/s. The kinetic energy of the car is:

KE=12×1200 kg×(20 m/s)2=12×1200×400=600×400=240,000 J=240 kJKE = \frac{1}{2} \times 1200 \text{ kg} \times (20 \text{ m/s})^2 = \frac{1}{2} \times 1200 \times 400 = 600 \times 400 = 240,000 \text{ J} = 240 \text{ kJ}

Example 4: A bullet of mass 0.02 kg is fired at a velocity of 300 m/s. The kinetic energy of the bullet is:

KE=12×0.02 kg×(300 m/s)2=12×0.02×90000=0.01×90000=900 JKE = \frac{1}{2} \times 0.02 \text{ kg} \times (300 \text{ m/s})^2 = \frac{1}{2} \times 0.02 \times 90000 = 0.01 \times 90000 = 900 \text{ J}

Gravitational Potential Energy

Gravitational potential energy (GPEGPE) is the energy an object has due to its position in a gravitational field. It is calculated as:

GPE=mghGPE = mgh

where:

  • GPEGPE is the gravitational potential energy (in joules, J)
  • mm is the mass of the object (in kilograms, kg)
  • gg is the acceleration due to gravity (approximately 9.8 m/s² on Earth)
  • hh is the height of the object above a reference point (in meters, m)

Example 5: A bag of rice with mass 25 kg is lifted to a shelf 2.5 m high. The gravitational potential energy gained is:

GPE=25 kg×9.8 m/s2×2.5 m=25×9.8×2.5=245×2.5=612.5 JGPE = 25 \text{ kg} \times 9.8 \text{ m/s}^2 \times 2.5 \text{ m} = 25 \times 9.8 \times 2.5 = 245 \times 2.5 = 612.5 \text{ J}

Example 6: A worker lifts a 40 kg load to a height of 1.8 m. The GPE gained by the load is:

GPE=40 kg×9.8 m/s2×1.8 m=40×9.8×1.8=392×1.8=705.6 JGPE = 40 \text{ kg} \times 9.8 \text{ m/s}^2 \times 1.8 \text{ m} = 40 \times 9.8 \times 1.8 = 392 \times 1.8 = 705.6 \text{ J}

Challenging Combined Questions

Example 7: A 2 kg object is dropped from a height of 10 m.

  • (a) Calculate the GPE at the top:
    GPE=mgh=2 kg×9.8 m/s2×10 m=2×9.8×10=196 JGPE = mgh = 2 \text{ kg} \times 9.8 \text{ m/s}^2 \times 10 \text{ m} = 2 \times 9.8 \times 10 = 196 \text{ J}
  • (b) Assuming no energy loss, its kinetic energy just before hitting the ground will be equal to the initial GPE:
    KE=196 JKE = 196 \text{ J}

Example 8: A cyclist (total mass 70 kg) moves at a speed of 5 m/s up a hill 4 m high.

  • (a) Calculate the initial kinetic energy:
    KE=12mv2=12×70 kg×(5 m/s)2=12×70×25=35×25=875 JKE = \frac{1}{2}mv^2 = \frac{1}{2} \times 70 \text{ kg} \times (5 \text{ m/s})^2 = \frac{1}{2} \times 70 \times 25 = 35 \times 25 = 875 \text{ J}
  • (b) Calculate the GPE gained at the top:
    GPE=mgh=70 kg×9.8 m/s2×4 m=70×9.8×4=686×4=2744 JGPE = mgh = 70 \text{ kg} \times 9.8 \text{ m/s}^2 \times 4 \text{ m} = 70 \times 9.8 \times 4 = 686 \times 4 = 2744 \text{ J}
  • (c) The total energy needed to reach the top is the sum of the initial kinetic energy and the GPE gained:
    Etotal=KE+GPE=875 J+2744 J=3619 JE_{\text{total}} = KE + GPE = 875 \text{ J} + 2744 \text{ J} = 3619 \text{ J}

Crane Lifting a Load

Example 9: A crane lifts a 500 kg load to a height of 20 m in 10 seconds. It then drops the load, which hits the ground with a speed of 19.8 m/s.

  • (a) Calculate the work done by the crane:
    W=mgh=500 kg×9.8 m/s2×20 m=500×9.8×20=4900×20=98000 J=98 kJW = mgh = 500 \text{ kg} \times 9.8 \text{ m/s}^2 \times 20 \text{ m} = 500 \times 9.8 \times 20 = 4900 \times 20 = 98000 \text{ J} = 98 \text{ kJ}

  • (b) Calculate the kinetic energy just before impact:
    KE=12mv2=12×500 kg×(19.8 m/s)2=12×500×392.04=250×392.04=98010 J=98.01 kJKE = \frac{1}{2}mv^2 = \frac{1}{2} \times 500 \text{ kg} \times (19.8 \text{ m/s})^2 = \frac{1}{2} \times 500 \times 392.04 = 250 \times 392.04 = 98010 \text{ J} = 98.01 \text{ kJ}

  • (c) Comment on the energy transformation in this scenario:

    The crane does work to lift the load, converting electrical energy into gravitational potential energy. When the load is dropped, the gravitational potential energy is converted into kinetic energy. The kinetic energy just before impact is slightly different from the work done by the crane due to energy losses (e.g., air resistance).