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A block with mass m = 17 kg rests on a frictionless table and is accelerated by

ID: 1320868 • Letter: A

Question

A block with mass m = 17 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4058 N/m after being compressed a distance x1 = 0.506 m from the spring

A block with mass m = 17 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4058 N/m after being compressed a distance x1 = 0.506 m from the spring??s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.1 m long. For this rough path, the coefficient of friction is ?k = 0.42. 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 = 0.129 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 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

1) How much work is done by the spring as it accelerates the block
- i did 1/2kx^2 and got 519

2) Work donr by spring = 0.5 * 4058 * 0.506^2 = 519.49 joules

3)What is the speed of the block right after it leaves the spring
0.5 * 17 * v^2 = 0.5 * 4058* 0.506^2
v = (4058 * 0.506^2) / 17)^0.5 =

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

Friction work = u * m * g * d = (0.42 * 17 * 9.8 * 2.1) = 146.94joules

please post other questions separately.

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