A block with mass m = 12 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 12 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4646 N/m after being compressed a distance x1 = 0.461 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.8 m long. For this rough path, the coefficient of friction is k = 0.41.

5. Instead, the spring is only compressed a distance x2 = 0.123 m before being released. How far into the rough path does the block slide before coming to rest?

6. What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?

7. If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is:

the same

three times greater

three times less

nine times greater

nine times less?

Explanation / Answer

5)

If the spring is compressed 0.123 m, the work done by the spring is 0.5 * 4646 * 0.123^2. So the initial KF of the block = 0.5 * 4646 * 0.123^2. When the block comes to rest, the KE = 0.
The work done by the friction force must decrease the kinetic energy of the block from (0.5 * 4646 * 0.123^2.) to 0.

0.41* 12 * 9.81 * ( length of rough path) = (0.5 * 4646 * 0.123^2.)
( length of rough path) = (0.5 * 4646 * 0.123^2.) / (0.41* 12 * 9.81) = 0.728 m

6) The work done by the spring must be "barely" greater than than the worjk done by the friction!

0.5 * k * x^2 = u * m* g * d
x^2 = (u * m* g * d) / (0.5 * k) = (0.41 * 12 * 9.81 * 0.728) / (0.5 * 4646) = 0.015129
x = 0.1253 m

If the the spring is compressed barely more than 0.123 meter, the block will come to rest at a point barely past the 0.728 m long rough patch!

If the rough patch is the original 2.8 m long:
x^2 = (u * m* g * d) / (0.5 * k) = (0.41 * 12 * 9.81 * 2.8) / (0.5 * 4646) =
x = 0.2412 m

7) W = 0.5*k*x^2

   W = 0.5*k*(3x)^2

W' = 9W

nine times greater

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