NUMERICAL RESPONSE QUESTIONS

Solve the following:

1. Calculate the work done in pushing a box with 150 N through a distance of 5 m.
(Ans. 750 J)

2. A boy weighing 75 N jumps up and gains 300 J of gravitational potential energy. Find the height to which the boy will rise.
(Ans. 4 m)

3. A 5 kg steel ball is dropped from the top of a 12 m high tower. What is the kinetic energy of the ball on hitting the ground? With what velocity will it hit the ground? (Neglect air resistance).
(Ans. 588 J and 15.3 m/s)

4. A 2.0 kg rock is dropped from a 20 m tall building. What are its kinetic and gravitational potential energies when the rock has fallen 15 m?
(Ans. Eₖ = 294 J and Eₚ = 98 J)

5. A rocket with a mass of 800 g is launched vertically upward with an initial speed of 30 m/s. (a) Assuming no air resistance, calculate the maximum height the rocket would reach. (b) If, due to air friction, the rocket only rises to 25 m, determine the work done against air resistance.
(Ans. (a) 45.92 m, (b) -164 J)

6. A 2 hp electric motor gives energy to a system that lifts a load of 100 kg to a height of 10 m in 1.5 s. Calculate (a) Input (work done by the motor on the system), (b) Output (load lifted by the system), and (c) Efficiency of the system.
(Ans. (a) 2238 J, (b) 980 J, (c) 22.8%)

What horsepower (hp) is required to pump up 2500 kg of water to a height of 100 m in 5 minutes?
(Ans. 95 hp)

  1. Work done in pushing a box:

    W=F×d

    Where:

    • F=150 N
    • d=5 m

    W=150 N×5 m=750 J

    Answer: 750 J


  1. Height to which the boy will rise: The potential energy gained is equal to the gravitational potential energy at the height .

    Ep=mgh

    Given:

    • Ep=300 J
    • m=75/9.8 kg≈7.65 kg
    • g=9.8 m/s2

    Solving for :

    h=Ep/mg=300 J/7.65 kg×9.8 m/s2≈4 m

    Answer: 4 m


  1. Kinetic energy and velocity of the ball on hitting the ground: The potential energy at the height will convert into kinetic energy at the ground.

    Ep=mgh

    Where:

    • m=5 kg
    • h=12 m
    • g=9.8 m/s2

    Ep=5 kg×9.8 m/s2×12 m=588 J

    The kinetic energy on hitting the ground is the same:

    Ek=588 J

    The velocity can be found from the kinetic energy formula:

    Ek=1/2mv2

    Solving for :

    v=√2Ek/m=2×588/5=15.3 m/s

    Answer: Kinetic energy = 588 J, Velocity = 15.3 m/s


  1. Kinetic and gravitational potential energies when the rock has fallen 15 m: Total height = 20 m, fallen height = 15 m, so the remaining height is 5 m.

    Gravitational potential energy when the rock has fallen 15 m:

    Ep=mgh

    Where:

    • m=2.0 kg
    • g=9.8 m/s2
    • h=5 m

    Ep=2.0 kg×9.8 m/s2×5 m=98 J

    The kinetic energy is the total potential energy lost:

    Ek=Total potential energy−Remaining potential energy

    The initial potential energy at height 20 m:

    Ep=2.0 kg×9.8 m/s2×20 m=392 J

    Therefore:

    Ek=392 J−98 J=294 J

    Answer: Kinetic energy = 294 J, Gravitational potential energy = 98 J


  1. Maximum height of the rocket: (a) The initial kinetic energy is converted into gravitational potential energy at maximum height:

    Kinetic energy=Gravitational potential energy

    Using:

    1/2mv02=mgh

    Where:

    • m=0.8 kg
    • v0=30 m/s
    • g=9.8 m/s2

    Solving for :

    h=v02/2g=302/2×9.8=45.92 m

    (b) The work done against air resistance is the difference in gravitational potential energy at the maximum height with and without air resistance:

    Wair resistance=mghwith air−mghwithout air           =0.8×9.8×25−0.8×9.8×45.92=−164 J

    Answer: (a) 45.92 m, (b) -164 J


  1. Efficiency of the system:

    (a) Input work done by the motor: The motor’s power output:

    P=Input work/Time

    Where:

    • Power = 2 hp = 2 \times 746 W = 1492 W
    • Time = 1.5 s

    Input work=P×Time=1492 W×1.5 s=2238 J

    (b) Output work (load lifted): The gravitational potential energy of the 100 kg load lifted:

    Ep=mgh=100×9.8×10=980 J

    (c) Efficiency:

    Efficiency=Output work/Input work×100           =980 J/2238 J×100               ≈22.8%

    Answer: (a) 2238 J, (b) 980 J, (c) 22.8%


  1. Horsepower required to pump water: Work done to lift 2500 kg of water:

    Ep=mgh

    Where:

    • m=2500 kg
    • h=100 m
    • g=9.8 m/s2

    Ep=2500×9.8×100=2,450,000 J

    Power required:

    P=Ep/Time=2,450,000 J/5×60=8166.67 W

    Converting to horsepower:

    Horsepower=8166.67/746≈10.95 hp

    Answer: 95 hp