A) \[mg(h+d)+\frac{1}{2}k{{d}^{2}}\]
B) \[mg(h+d)-\frac{1}{2}k{{d}^{2}}\]
C) \[mg(h-d)-\frac{1}{2}k{{d}^{2}}\]
D) \[mg(h-d)+\frac{1}{2}k{{d}^{2}}\]
Correct Answer: B
Solution :
Key Idea Work done is equal to change in energy of body. Situation is shown in figure. When mass \[m\] falls vertically on spring, then spring is compressed by distance \[d\]. Hence, net .work done in the process is \[W=\]Potential energy stored in the spring + Loss of potential energy of mass \[=mg(h+d)-\frac{1}{2}k{{d}^{2}}\]You need to login to perform this action.
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