206k views
1 vote
A 25 kg box on a horizontal frictionless surface is moving to the right at a speed of +2.0 m/s. The box hits and remains attached to one end of a spring of negligible mass whose other end is attached to a wall. As a result, the spring compresses a maximum distance of 0.50 m, and the box then sticks to the spring and oscillates back and forth.

a. The spring does work on the box from the moment the box first hits the spring to the moment the spring first reaches its maximum compression. Calculate the magnitude of the work done by the spring.

b. Calculate the spring constant of the spring.

c. Calculate the frequency of the oscillation of the box.

d. Let x = 0 be the point where the box makes contact with the spring, with positive x directed toward the right. On the axes below, sketch the velocity a of the oscillating box as a function of position x for the range x = -0.50 m to x = +0.50 m.

User Palani
by
8.1k points

1 Answer

2 votes

a. The work done by the spring on the box is equal to the change in kinetic energy of the box as it comes to a stop and then compresses the spring. The initial kinetic energy of the box is:

KEi = (1/2)mv² = (1/2)(25 kg)(2.0 m/s)² = 50 J

At maximum compression, all of the kinetic energy of the box has been converted to potential energy stored in the compressed spring. Therefore, the work done by the spring is:

W = KEi = 50 J

b. The potential energy stored in a spring is given by:

PE = (1/2)kx²

where k is the spring constant and x is the displacement from the equilibrium position. At maximum compression, x = -0.50 m and the potential energy stored in the spring is equal to the initial kinetic energy of the box:

PE = KEi = 50 J

Substituting the given values:

(1/2)k(-0.50 m)² = 50 J

Solving for k:

k = (2)(50 J) / (0.50 m)² = 800 N/m

c. The frequency of oscillation of the box is given by:

f = (1/2π)√(k/m)

Substituting the given values:

f = (1/2π)√(800 N/m / 25 kg) = 2.54 Hz

d. The velocity of the box as a function of position x can be sketched as follows:

At the equilibrium position (x = 0), the box has zero velocity. As it moves to the right, it begins to compress the spring, which slows it down until it reaches maximum compression at x = -0.50 m. At this point, the box has zero velocity again and begins to move back to the right. As it moves away from the spring, it gains speed until it reaches the equilibrium position, where it has maximum velocity. It continues to move to the right and compress the spring again, repeating the cycle. The graph of velocity versus position is a sinusoidal wave, with the maximum velocity at x = 0 and the minimum velocity at x = ±0.50 m.

User Rob Wright
by
7.7k points

No related questions found