Examkin

Question

Difficulty: HardWork, Energy and Power

A box of mass 2 kg2\text{ kg} slides down a rough inclined plane from a height of 5 m5\text{ m}. If it reaches the bottom of the incline with a speed of 6 m s−16\text{ m s}^{-1}, what is the work done against friction during the descent? (Take g=10 m s−2g = 10\text{ m s}^{-2})

  1. 64 J64\text{ J}Answer
  2. B
    36 J36\text{ J}
  3. C
    100 J100\text{ J}
  4. D
    136 J136\text{ J}

Answer

64 J64\text{ J}
According to the principle of conservation of energy, the work done against friction equals the loss in total mechanical energy. The initial potential energy is mgh=2×10×5=100 Jmgh = 2 \times 10 \times 5 = 100\text{ J}, and the final kinetic energy is 12mv2=12×2×62=36 J\frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times 6^2 = 36\text{ J}. Subtracting final kinetic energy from initial potential energy yields 100 J−36 J=64 J100\text{ J} - 36\text{ J} = 64\text{ J}.

Step-by-Step Solution

1
Calculate the initial potential energy (EpE_p) at height h=5 mh = 5\text{ m}.
Ep=mgh=2 kg×10 m s−2×5 m=100 JE_p = mgh = 2\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 100\text{ J}
At the top of the incline, all mechanical energy is stored as gravitational potential energy.
2
Calculate the final kinetic energy (EkE_k) at the bottom where v=6 m s−1v = 6\text{ m s}^{-1}.
Ek=12mv2=12×2 kg×(6 m s−1)2=36 JE_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 2\text{ kg} \times (6\text{ m s}^{-1})^2 = 36\text{ J}
At the bottom of the incline, the remaining mechanical energy is kinetic energy.
3
Apply the work-energy theorem to find the work done against friction (WfW_f).
Wf=Ep−Ek=100 J−36 J=64 JW_f = E_p - E_k = 100\text{ J} - 36\text{ J} = 64\text{ J}
The non-conservative friction force reduces the total mechanical energy by doing work against motion.

Key Concept

Work-Energy Theorem and Conservation of Energy with Friction
Rate this question

Topics

Work, Energy and Power

More questions from this topic