A 25-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 12 J. What is the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm?
A) −4.0 J
B) −3.5 J
C) −2.4 J
D) −2.9 J
E) −15 J

Question

Quizplus · Accepted Answer

First, we need to find the potential energy stored in the spring when the block is pulled 10 cm to the right from its equilibrium position:

$U_s=\frac{1}{2}kx^2=\frac{1}{2}(8.0	imes 10^3	ext{ N/m})(0.1	ext{ m})^2=4	ext{ J}$

When the block has moved 2.0 cm toward its equilibrium position, its displacement from the equilibrium position is:

$x=0.1 	ext{ m} - 0.02 	ext{ m} = 0.08 	ext{ m}$

The kinetic energy of the block at this point is given as 12 J. Therefore, we have:

$K=\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{\mathrm{d}x}{\mathrm{d}t}ight)^2$

$v^2=\frac{2K}{m}=\frac{2(12	ext{ J})}{25	ext{ kg}}=0.96	ext{ m}^2/	ext{s}^2$

$\frac{\mathrm{d}x}{\mathrm{d}t}=v=\sqrt{0.96}=0.98 	ext{ m/s}$

The force exerted by the spring on the block when it is at displacement $x$ from the equilibrium position is given by:

$F=-kx$

The work done by this force during the block's motion from 10 cm to 8 cm from the equilibrium position is:

$W=\int_{0.1	ext{ m}}^{0.08	ext{ m}} -kx \mathrm{d}x=-\frac{1}{2}kx^2\Big|_{0.1	ext{ m}}^{0.08	ext{ m}} \approx -0.5	ext{ J}$

The work done by the frictional force is equal to the change in mechanical energy, which is:

$\Delta E = W_	ext{friction}= K_f - K_i - W_s$

where $K_i$ is the initial kinetic energy of the block when it is released, $W_s$ is the work done by the spring force, and $K_f$ is the final kinetic energy when the block is at 8 cm from the equilibrium position. $K_i$ and $W_s$ are both 0, since the block is released from rest, and the work done by the spring force is equal to the potential energy stored in the spring at the initial displacement.

$\Delta E = K_f -  W_s=-W_	ext{friction}$

$\Delta E = \frac{1}{2}mv_f^2 - \frac{1}{2}kx^2 \approx \frac{1}{2}(25	ext{ kg})(0.98	ext{ m/s})^2 - 4	ext{ J} \approx -2.4	ext{ J}$

Therefore, the change in mechanical energy caused by the frictional force on the block as it moves the 2.0 cm is -2.4 J, which is answer choice C.

A 25-Kg Block on a Horizontal Surface Is Attached to a Light