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Chapter 3: Problem 68

cant copy graph A box of mass \(m\) is released from rest at the top of a frictionless ramp tilted at an angle \(\theta\) from the horizontal. The box slides down a distance \(s\) and collides with spring with a spring constant \(k\). The spring then compresses a distance \(x\). To find \(x\) one should (A) equate \(m g s\) to \(1 / 2 k^2\) and solve for \(x\). (B) equate \(m g s \cos \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (C) equate \(m g s \sin \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (D) equate \(m g(x+s) \sin \theta\) to \(1 / 2 k x^2\) and solve for \(x\). (E) equate \(m g\) to \(k x\) and solve for \(x\).

Short Answer

Expert verified
Answer: The correct equation is (C) \(mgs\sin \theta = \frac{1}{2}kx^2\), where \(m\) is the mass of the box, \(g\) is the acceleration due to gravity, \(s\) is the distance along the ramp, \(\theta\) is the angle of inclination of the ramp, \(k\) is the spring constant, and \(x\) is the compression of the spring.

Step by step solution

01

State work-energy theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy: \(W = \Delta KE\)
02

Apply work-energy theorem to the box

When the box is released from the top of the ramp and slides down a distance \(s\), it loses gravitational potential energy and gains kinetic energy. The work-energy theorem can be written as \(mgh = \frac{1}{2} mv^2\), where \(h\) is the vertical height and \(v\) is the final velocity of the box. We can rewrite this, taking into account the \(\theta\) and \(s\), \(mgs\sin \theta = \frac{1}{2} mv^2\), where \(v\) is the velocity at the bottom of the ramp.
03

Apply work-energy theorem to the spring

When the box collides with the spring, its kinetic energy is converted into potential energy stored in the spring. The work-energy theorem for the spring can be written as \(\frac{1}{2}mv^2 = \frac{1}{2}kx^2\).
04

Combine equations from Steps 2 and 3

Combining the equations from steps 2 and 3, we have the equation: \(mgs\sin \theta = \frac{1}{2}kx^2\).
05

Compare the combined equation with the options given

Comparing the combined equation with the given options, we see that option (C) equating \(m g s \sin \theta\) to \(\frac{1}{2} k x^2\) and solving for \(x\), matches our conclusion. Therefore, the correct answer is (C).

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