JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Work, Power & Energy

A ball of mass 100 g is dropped from a height

h = 10

cm on a platform fixed at the top of vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance

\frac{h}{2}

. The spring constant is ______

Nm^{- 1}

. (Use

g = 10 ms^{- 2}

)

Answer: 120

Solution

By conservation of energy, the gravitational potential energy lost equals the elastic potential energy stored in the spring.

The ball falls through a total distance of

h + \frac{h}{2} = \frac{3 h}{2}

, and the spring is compressed by

x = \frac{h}{2}

.

Using energy conservation:

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

m g (\frac{3 h}{2}) = \frac{1}{2} k (\frac{h}{2})^{2}

Substituting values:

m = 0.1

kg,

g = 10

m/s²,

h = 0.1

0.1 \times 10 \times \frac{3 \times 0.1}{2} = \frac{1}{2} k (0.05)^{2}

0.15 = \frac{1}{2} k \times 0.0025

k = \frac{0.15 \times 2}{0.0025} = \frac{0.3}{0.0025} = 120 N/m

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Work, Power & Energy

Solution

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