JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Work, Power & Energy

A block of mass '

m

' (as shown in figure) moving with kinetic energy

E

compresses a spring through a distance 25 cm when, its speed is halved. The value of spring constant of used spring will be

n E

^{- 1}

for

n

= _____.

Answer: 24

Solution

Using the work-energy theorem, the work done by the spring equals the change in kinetic energy.

Initial kinetic energy:

K_{i} = E = \frac{1}{2} m v^{2}

Final kinetic energy (speed halved):

K_{f} = \frac{1}{2} m (\frac{v}{2})^{2} = \frac{E}{4}

Work done by spring:

W = - \frac{1}{2} k x^{2}

(negative because it opposes motion)

Applying work-energy theorem:

- \frac{1}{2} k x^{2} = K_{f} - K_{i} = \frac{E}{4} - E = - \frac{3 E}{4}

\frac{1}{2} k x^{2} = \frac{3 E}{4}

k = \frac{3 E}{2 x ^{2}}

With

x = 25

= 0.25

= \frac{1}{4}

k = \frac{3 E}{2 \times ( 1/4 ) ^{2}} = \frac{3 E}{2 \times 1/16} = \frac{3 E \times 16}{2} = 24 E N/m

Therefore,

n = 24

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer