Work Energy and Power - Result Question 5

7. A vertical spring with force constant $k$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

[2007]

(a) $m g(h+d)-\frac{1}{2} k d^{2}$

(b) $m g(h-d)-\frac{1}{2} k d^{2}$

(c) $mg(h-d)+\frac{1}{2} kd^{2}$

(d) $mg(h+d)+\frac{1}{2} kd^{2}$

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Answer:

Correct Answer: 7. (a)

Solution:

$ =8+5 $

$ =13 J . $

. (a)

$W=$ Potential energy stored in the spring + Loss of potential energy of mass

$\Rightarrow W_a=-8.75 J$

which is the work done due to air resistance

Work done due to gravity $=mgh$

$=10^{-3} \times 10 \times 10^{3}=10 J$

$W=m g(h+d)-\frac{1}{2} k d^{2}$

Gravitational potential energy of ball gets converted into elastic potential energy of the spring.

$m g(h+d)=\frac{1}{2} k d^{2}$

Net work done $=mg(h+d)-\frac{1}{2} k d^{2}=0$

$\frac{W_P}{W_Q}=\frac{\frac{1}{2} k_p x^{2}}{\frac{1}{2} k_Q x^{2}}=\frac{k_p}{k_Q}$

As $k_p>k_Q$ so, $W_p>W_Q$