A 1.0-kg block is released from rest at the top of a frictionless incline that makes an angle of 37° with the horizontal. An unknown distance down the incline from the point of release, there is a spring with k = 200 N/m. It is observed that the mass is brought momentarily to rest after compressing the spring 0.20 m. How far does the mass slide from the point of release until it is brought momentarily to rest?
A) 0.98 m
B) 0.68 m
C) 0.82 m
D) 0.55 m
E) 0.20 m

Question

Quizplus · Accepted Answer

Using conservation of energy, the gravitational potential energy at the top is converted to spring potential energy at the bottom:

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

where h is the height of the incline and x is the compression of the spring.

Solving for x gives:

$x = \sqrt{\frac{2mgh}{k}} = \sqrt{\frac{2 \cdot 1.0 \cdot 9.8 \cdot \sin37}{200}} \approx 0.2m$

Therefore, the distance traveled after the spring is compressed is equal to the total distance traveled minus the distance the spring is compressed:

$d = \frac{h}{\sin 	heta} = \frac{1.0 \cdot 9.8}{\sin37} \approx 1.55m$

$d_{remaining} = d - x \approx 1.55m - 0.2m \approx 1.35m$

However, this is the distance traveled from the top of the incline, not the point of release. The distance traveled from the point of release can be found using trigonometry:

$d_{release} = d_{remaining} \cos 	heta = 1.35m \cos37 \approx 0.68m$

Therefore, the answer is B.

A 10-Kg Block Is Released from Rest at the Top of of a Frictionless