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Question

Part A

A 50 g ice cube on a 30 degree (to the horizontal) wooden slope is held against a spring, of spring constant 25 N/m. The spring is compressed 10 cm. Consider the coefficient of friction for ice on wood to be zero. When the ice cube is released and launched up the ramp, what total distance will it travel up the slope before reversing direction?

Part B

If the ice cube is replaced by a 50 g wooden cube, how far will it travel up the slope?

Answer 1

The potential energy in the spring is 0.5kx^2 = 0.5*25*(0.1)^2 = 0.125 J.

This energy will turn into gravitation potential energy at the maximum height.

GPE = mgh = 0.05*9.8*h = 1/8

h = 1/(8*9.8*0.05) = 0.255 cm.

h = 0.255 cm is the increase in altitude. Divide this by sin30 to get the distance up the slope.

0.255/sin30 = 51 cm.

When there is coefficient of friction = 0.2, the normal force is mgcos30 = 0.424 N.

Friction force = 0.2*0.424 = 0.08487 N.

The initial energy is converted to GPE plus work done by friction

0.125 = mgs(sin30) + Fs

s = 1/[8(mg(sin30) + F]

= 1/[8(0.245 + 0.08487] = 0.3789 cm.

I was going to be lazy and calculate the work done to go up 51 cm, the answer in the first part, and then proportion it down. I decided that might be hard to follow. I will use it as a check.

Work done to slide up 51 cm = 0.51*0.08487 = 0.0432837 J

51*(.125/(0.04328 + 0.125)) = 0.3789 cm. It checks.

These distances are measured from the compressed spring position.