A block with mass m = 15 kg rests on a frictionless table and is accelerated by

Question

A block with mass m = 15 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4568 N/m after being compressed a distance x1 = 0.471 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.2 m long. For this rough path, the coefficient of friction is k = 0.4.

1)How much work is done by the spring as it accelerates the block?

______________J

2)What is the speed of the block right after it leaves the spring?

___________m/s

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

______________J

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

_____________m/s

5)Instead, the spring is only compressed a distance x2 = 0.138 m before being released.

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

________________m

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?

____________m

Explanation / Answer

Part 1: Work done on the block
=====
F = kx
F = 4568N/m * 0.471 m
F = 2151.5 N

a = F/m = 2151.5/15 = 143.43 m/s^2
Energy = 1/2 kx^2 = 1/2 * 4568 * 0.471^2
Energy (Work) = 506.68 J

Part 2: Speed of the block after release.
=====
Kinetic energy of the block = 1/2 m v^2
506.68 = 1/2 * 15 kg * v^2
67.56 = v^2
v = 8.22 m/s

Part 3
=====
frictional_force = N*ik
frictional_force = 15*9.81*0.4
frictional_force = 58.86 N

Work = F * d
Work = 58.86 * 2.2 = 129.49 J

Part 4
=====
Energy before rough Spot - Energy used by friction = remaining kinetic energy.
Remaining KE = 506.68 - 129.49 = 377.19 J
Ke = 1/2 m*v^2
377.19 = 1/2 * 15 * v^2
v^2 = 50.292
v = 7.09 m/s

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