A 10.0 kg block is to be projected across a rough horizontal floor by means of a

Question

A 10.0 kg block is to be projected across a rough horizontal floor by means of a compressed spring device. The spring has a spring constant of 5000 N/m, while the coeffiecient of static and kinetic friction between the block and the floor are known to be 0.575 and 0.385, respectively. The spring is initially compressed a distance of 0.250 m. Answer the following:

(a) What is the potential energy, in joules, stored in the spring before it is released?

(b) What is the work done by friction, in joules, as the block slides to a stop?

(c) How far, in meters, will the block slide from its initial position against the compressed spring before it stops moving?

Explanation / Answer

A)it's potential energy PE = 1/2 kx^2 = 0.5*(0.250)^2 *( 5000)= 156.25 Joules.

B)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 %u2022 m %u2022 v^2, we know that:

1/2 %u2022 m %u2022 v^2 = U
m %u2022 v^2 = 2U
v^2 = 2U/m
v = SQRT (2U/m)

By substituting m = 10 kg and U = 156.25 Joules, you get:

v = SQRT (156.25 J / 10 kg)
v = SQRT (15.625 m^2/s^2)
v = 3.95 m/s

hence change in KE = work done by friction

C)

you can apply s=ut-(1/2)at^2

Hope this helps!

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