Q4.17 Dissipation (S)

A block of mass 10 kg slides a distance 1 m down a 45 slope. The coefficient of kinetic friction between the block and slope is 0.2. By how much does the kinetic energy of the block increase?
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Hint

How much does the potential energy of the block decrease?
 

Solution

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Solution

As in question Q4.16, the magnitude of the normal reaction exerted by the sloping surface on the block is

FN=mgcosθ.

The magnitude of the frictional force is

FF=μkmgcosθ

In travelling d=1m down the slope, the block drops a vertical distance of $h=1/\sqrt{2}=0.71\,m$. The gravitational potential energy thus changes by $mg/\sqrt{2}=69.3\,J$. In the absence of friction, all of this would appear as an increase in kinetic energy. But in 1 m of motion down the slope, the frictional force dissipates FFd=13.9 J of mechanical energy, leaving 55.4 J as kinetic energy. (Note that the frictional force is path dependent; a different path means a different answer.)