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A spring on a wooden ramp has a spring constant of 453 N/M. A small cube is to be launched and has a mass of 97 grams. Ramp angle from the horizontal is 52 degrees above the horizontal. For simplicity, assume that the part of the wooden ramp which is underneath the spring is highly polished and very slick; you may assume no friction on the cube by the ramp when the cube is moving on this portion of the ramp. For the rest of the wooden ramp, the coefficients of friction between the ramp surface and the cube surface are*0.59*for static friction and*0.37*for kinetic friction. Measured from the equilibrium position of the free end of the mounted spring, the distance to the top of the ramp is*21*cm (this is measured along the ramp). The spring for this question is compressed 5.6cm.

What is the work done on the cube by gravity, from the instant just after the trigger is released to the instant just before the sliding cube leaves the ramp?

2. Homework Equations

Net Work = Force dot product delta r = Change in Kinetic energy

Hooke's Law

3. The Attempt at a Solution

Mass times Gravity times delta r, which I chose as 0.266 meters because my spring is compressed 0.056 meters + my displacement from the spring equilibrium to the end of the ramp is 0.21m.

I also tried (Mass time Gravity)sin52 degrees times delta r.

Help is very much appreciated thank you.