Question

A 1.0 kg block collides with a horizontal light spring of force constant 2 $\frac{N}{m}$. The block compresses the spring 4m from the rest position. Assuming that the coefficient of kinetic friction between the block and the horizontal surface is 0.25, what was the speed of the block at the instant of collision?

• A. 52
• B.
• C. 25
• D. none of these

Answer 1

The correct option is B

So Let's see what is happening

K= 2 $\frac{N}{m}$

X = 4m

M= 0.25

Assume block hits with velocity V $\frac{m}{s}$

The spring will compress till the point the block stops.

Also friction is acting all this while.

Applying work energy theorem.

(Only friction and spring forces are acting)

$W_{fr}+W_{sp}=-\frac{1}{2}m{v}^2-\mu mgx-\frac{1}{2}k{x}^2=-\frac{1}{2}m{v}^2\mu mgx+\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2\frac{1}{4}\left(1\right)\left(10\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(16\right)=\frac{1}{2}\left(1\right)v^{2}10+16=\frac{v^2}{2}v=\sqrt{26\times 2}=\sqrt{52}m/s$