Aforce of 1,500 n is moved through 0.10 m in order to compress a steel cylinder. the cylinder was compressed 0.010 m by a force of 13,000 n. the efficiency of the compression machine is what % ?

115
0.87
87
8.7

We first calculate the work effectively done by the force when it compressed the steel cylinder. This is solved by multiplying the distance that the cylinder was compressed by and its causing force.
0.010 m * 13,000 N = 130 N-m

Then, we solve for the work done by the 1,500 N force:
0.10 m * 1,500 N = 150 N-m

We then take the ratio of the efficient:experimental
130 N-m / 150 N-m = 0.87 = 87%

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.