A 20-kg block on a horizontal surface is attached to a light spring (force constant = 8.0 kN/m).The block is pulled 10 cm to the right from its equilibrium position and released from rest.When the block has moved 2.0 cm toward its equilibrium position,its kinetic energy is 13 J.How much work is done by the frictional force on the block as it moves the 2.0 cm?
A)-2.5 J
B)-1.4 J
C)-3.0 J
D)-1.9 J
E)-14 J

Question

Quizplus · Accepted Answer

The potential energy of the spring is given by:
$U=\frac{1}{2}kx^2$
where k is the force constant of the spring and x is the displacement from equilibrium.

At the starting point, the block has a potential energy of:
$U_i=\frac{1}{2}(8.0	imes 10^3 N/m)(0.1 m)^2=40 J$

At the point where the block has moved 2.0 cm toward its equilibrium position, its potential energy is:
$U_f=\frac{1}{2}(8.0	imes 10^3 N/m)(0.08 m)^2=25.6 J$

The work done by the spring force is:
$W_s=U_i-U_f=14.4 J$

Using the work-energy principle:
$W_s+W_{fric}=\Delta K$
where $W_{fric}$ is the work done by the frictional force and $\Delta K$ is the change in kinetic energy.

Solving for $W_{fric}$ gives:
$W_{fric}=\Delta K-W_s=13 J -14.4 J=-1.4 J$

The negative sign indicates that the frictional force is in the opposite direction to the motion of the block. 
Thus, the best choice is B.

A 20-Kg Block on a Horizontal Surface Is Attached to a Light