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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Work and Energy

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Work and Energy

A block of mass 30.0 kg is lowered by a chain. Determine the work...

A block of mass 30.0 kg30.0 \text{ kg} is lowered by a chain. Determine the work done by the chain if the block acquires a speed of 40.0 cm/s40.0 \text{ cm/s} after dropping down a distance of 2.00 m2.00 \text{ m}. (Take g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Options

  1. A−585.6 J-585.6 \text{ J}
  2. B−590.4 J-590.4 \text{ J}
  3. C−588.0 J-588.0 \text{ J}
  4. D585.6 J585.6 \text{ J}

Correct answer

A. −585.6 J-585.6 \text{ J}

Step-by-step solution

Given m=30.0 kgm = 30.0 \text{ kg}, h=2.00 mh = 2.00 \text{ m}, v=40.0 cm/s=0.40 m/sv = 40.0 \text{ cm/s} = 0.40 \text{ m/s}, and u=0u = 0.

According to the work-energy theorem, the net work done on the block is equal to its change in kinetic energy:

Wnet=ΔKW_{\text{net}} = \Delta K

Wgravity+Wchain=12mv2−12mu2W_{\text{gravity}} + W_{\text{chain}} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2

The work done by gravity is:

Wgravity=mgh=30.0×9.8×2.00=588.0 JW_{\text{gravity}} = mgh = 30.0 \times 9.8 \times 2.00 = 588.0 \text{ J}

The change in kinetic energy is:

ΔK=12×30.0×(0.40)2=15.0×0.16=2.4 J\Delta K = \dfrac{1}{2} \times 30.0 \times (0.40)^2 = 15.0 \times 0.16 = 2.4 \text{ J}

Substituting these values into the work-energy equation:

588.0+Wchain=2.4588.0 + W_{\text{chain}} = 2.4

Wchain=2.4−588.0=−585.6 JW_{\text{chain}} = 2.4 - 588.0 = -585.6 \text{ J}
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