A spring with a spring constant of 38 N/m is placed on a horizontal smooth surfa

Question

A spring with a spring constant of 38 N/m is placed on a horizontal smooth surface and a mass of 3.3 kg is pushed against it. As a result the spring is compressed by 30 cm. The mass is then released. A) What is the amount of potential energy stored in the spring when it is compressed? B) What is the kinetic energy of the mass after the spring is released and the mass is no longer in contact with the spring? C) What is the velocity of the mass as it leaves contact with the spring? D) If the spring is now placed vertically with the mass on top, how much is the spring compressed?

Explanation / Answer

The potential energy stored in a spring is given by this formula:

U = 1/2 • k • x^2

Where U is the potential energy, k is the spring constant, and x is the distance that the spring is compressed or stretched. Thus, in this case:

U = 1/2 • k • x^2
U = 1/2 • 38 N/m • (0.3 m)^2
U = 1.53 Joules

The kinetic energy of the block will be equal to the potential energy of the spring. This happens because all of the energy stored as elastic potential energy in the spring is converted into kinetic energy when the spring is released. Since KE = 1/2 • m • v^2, we know that:

1/2 • m • v^2 = U
m • v^2 = 2U
v^2 = 2U/m
v = SQRT (2U/m)

By substituting m = 3.3 kg and U = 1.53 Joules, you get:

v = SQRT (1.53*2/3.3)
v = 0.96 m/s

d) mg = kx

3.3*10= 38x

x=33/38

=0.868 mts

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