You are pushing a 60 kg crate up a plane inclined at 25 degrees with respect to

Question

You are pushing a 60 kg crate up a plane inclined at 25 degrees with respect to the horizontal direction. The coefficient of kinetic friction between the surfaces of the crate and the inclined plane is Mu_k = 0.17. Assume that you are applying a constant push parallel to the plane. Now suppose that you also know that over a certain segment that is 1.6 m long, the velocity of the cart was 1.5 m/s and 0.5 m/s at the beginning and end points of the segment. How much of your energy did you transfer to the cart while pushing on the cart over the above segment?

Explanation / Answer

Let us consider the initial point as our reference point , therefore the potential energy at this point is zero.
And the kinetic energy = (1/2)mV2 = (1/2)*(60)*(1.52 ) = 67.5 J
So total energy at the beginning = 67.5 J
Now at the end point after covering 1.6 m along the plane.
The increase in the height of the block
= 1.6Sin25 = 0.676 m
Therefore the increase in the Potential energy = mgH = 60*9.81*(0.676) = 397.8936 J
Kinetic energy at this point = (1/2)*(60)(0.52 ) = 7.5 J
Therefore total energy = 397.8936 + 7.5 = 405.39 J
Change in energy = 405.39 - 67.5 = 337.896 J

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