One end of a spring is fixed to a wall and the other end is connected to a block of mass
1kg. Coefficient of friction between the block and the floor is $μ = 0.2$ . The block is pushed to compress the spring by 10 cm and released. Force constant of the spring is $k = 50 N . m^{- 1}$ . The block finally comes to rest. $(g = 10 m / s^{2})$

Which of the following is true?

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Finally, the block must comes to rest at the position where the spring is relaxed.

The spring will get relaxed more than once before the block finally stops.

The spring will get relaxed only once before the block finally stops.

Total distance travelled by the block before it finally comes to rest is 12 cm.

answer is D, C.

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Detailed Solution

The block will finally come to rest at a position where the spring force is less than maximum friction on the block.
   $k x \leq μ m g \Rightarrow 50 x \leq 0.2 \times 1 \times 10$
$\Rightarrow x \leq 0.04 m \Rightarrow x \leq 4 c m$
When the block reaches the relaxed position of the spring for the first time, its kinetic energy will be
$K E_{0} = \frac{1}{2} k (x_{0}^{2}) - μ m g . x_{0}$ where $x_{0}$ = 10 cm = 0.1 m
   $= \frac{1}{2} \times 50 \times {(0.1)}^{2} - 2 \times 0.1 = 0.05 J$
Let it move further by x (stretching the spring) before it comes to rest.
$0.05 J = \frac{1}{2} k x^{2} + μ m g . x$
$\Rightarrow 0.05 = 25 x^{2} + 2 x$ $\Rightarrow 25 x^{2} + 2 x - 0.05 = 0$
   $\Rightarrow x = \frac{- 2 \pm \sqrt{4 + 5}}{50} = \frac{- 2 \pm 3}{50}$ $= \frac{1}{50} m = 2 c m$
The block has come to rest when $x \leq 4 c m$ . Therefore, it will not move further.
Total distance travelled = 10 cm + 2 m = 12 cm.