JEE MainPhysicsModerateWork Power & Energy

A block of mass 60 kg is released from rest when compression in
the spring is 2 m (natural length of spring is 8 m). Surface AB is
smooth while surface BC is rough. Block travels x distance before coming
to complete rest. Value of x is: $\lbrack g = 10m/s^{2}\rbrack$

Show answer and solution

By Work Energy
Theorem

W_{net} = \Delta K.E = 0 \Rightarrow \frac{1}{2}k(x_{0}^{2} - x^{2}) - \mu mgx = 0

\Rightarrow \frac{1}{2} \times 200(2^{2} - x^{2}) = \frac{1}{2} \times 60 \times 10x

\Rightarrow x = 1m

Also at this moment

f_{\max} > kx

So, block will not move so total distance travelled =2+1=3 m.

Source: Grass-G / P-16 Physics paper 25 Nov First.docx

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