A block with mass m = 14 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 14 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4154 N/m after being compressed a distance x1 = 0.51 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.7 m long. For this rough path, the coefficient of friction is k = 0.44.

1) How much work is done by friction as the block crosses the rough spot?

J

2) What is the speed of the block after it passes the rough spot?

m/s

Instead, the spring is only compressed a distance x2 = 0.118 m before being released.

3)How far into the rough path does the block slide before coming to rest?

m

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

m

5) 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

Part
=====
frictional_force = N*ik
frictional_force = 14*9.81*0.44
frictional_force = 60.3 N

Work = F * d
Work = 60.3 * 2.7 = 162.9 J

Part : Work done on the block
=====
F = kx
F = 4154N/m * 0.51 m
F = 2118.5 N

a = F/m = 2118.5/14 = 151.3 m/s^2
Energy = 1/2 kx^2 = 1/2 * 4154 * 0.51^2
Energy (Work) = 540.2 J

Part : Speed of the block after release.
=====
Kinetic energy of the block = 1/2 m v^2
540.2 = 1/2 * 14 kg * v^2
v = 8.78 m/s

Part
=====
Energy before rough Spot - Energy used by friction = remaining kinetic energy.
Remaining KE = 540.2 - 162.9 = 377.3 J
Ke = 1/2 m*v^2
377.3 = 1/2 * 14 * v^2
v = 7.34 m/s

=====

PART:

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 9 times the original work.

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