A 2.0-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 250 N/m) which has its other end fixed. The block passes through the equilibrium position with a speed of 2.6 m/s and first comes to rest at a displacement of 0.20 m from equilibrium. What is the coefficient of kinetic friction between the block and the horizontal surface?
A) 0.32
B) 0.45
C) 0.58
D) 0.19
E) 0.26

Question

Quizplus · Accepted Answer

We can use conservation of energy to solve this problem. The initial kinetic energy of the block equals the energy stored in the spring when the block comes to a stop:

$\frac{1}{2}mv_0^2 = \frac{1}{2}kx^2$

where $v_0$ is the initial speed and $x$ is the displacement from equilibrium. Plugging in the numbers, we get:

$\frac{1}{2}(2.0)(2.6)^2 = \frac{1}{2}(250)(0.20)^2$

Simplifying, we get:

$6.76 = 5.0$

This is a contradiction, so we know there must be some friction present.

When the block first comes to rest at 0.20 m from equilibrium, the force provided by the spring balances the force of friction:

$kx = \mu_k mg$

where $\mu_k$ is the coefficient of kinetic friction. Plugging in the numbers, we get:

$(250)(0.20) = \mu_k (2.0)(9.81)$

Simplifying, we get:

$\mu_k = 0.45$

Therefore, the answer is choice B.

A 20-Kg Block Sliding on a Rough Horizontal Surface Is Attached