Question 13: Units , Dimensions and Error Analysis

Question

A vertical spring of spring constant $24 \mathrm{Nm}^{-1}$ is fixed on a table. A ball of mass 0.5 kg at a height 2 m above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance of 50 cm . The net work done in the process is:

Accepted Answer

LetÃ¢ÂÂs track the work done by gravity and by the spring as the ball falls 2 m and then compresses the spring by 0.5 m.
Height drop:
Total vertical displacement = 2.0 m + 0.5 m = 2.5 m
Work done by gravity:
$$W_g = m g \,\Delta h = (0.5\,	ext{kg})(10\,	ext{m/s}^2)(2.5\,	ext{m}) = 12.5\ 	ext{J}$$
Work done by the spring (opposes motion):
$$W_s = -	frac12\,k\,x^2             = -	frac12\,(24\ 	ext{N/m})\,(0.50\ 	ext{m})^2             = -3.0\ 	ext{J}$$
Net work on the ball:
$$W_{m net} = W_g + W_s = 12.5\ 	ext{J} - 3.0\ 	ext{J} = 9.5\ 	ext{J}$$
Answer: Option D (9.5 J).

Answer

6.5 J

Answer

10.5 J

Answer

12.5 J

Answer

9.5 J

Units , Dimensions and Error Analysis

A vertical spring of spring constant 24 \mathrm{Nm}^{-1} is fixed on a table. A ball of mass 0.5 kg at a height 2 m above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance of 50 cm . The net work done in the process is:

Choose the correct answer: