The force F acting in a constant direction on the 20kg block

The force F, acting in a constant direction on the 20-kg block, has a magnitude which varies with positions of the block. Determine the speed of the block after it slides 4m. When s = 0 the block is moving to the right at 2 m/s. The coefficient of kinetic friction between the block and surface is mu_k = 0.1.

Solution

Here, we have a block sliding up an incline with coefficient of kinetic friction Uk = 0.1. The block is moving with the speed of 2 m/s. We will determine the net work done on the block by the time it moves by 1 m and then use the principle of conservation of energy to find the block\'s speed.

Now, at any point s away from the start, force acting on the block is given as F = 50s²

This force acts at an angle to the block and hence will also contribute to the normal reaction of the block.

N = 3F/5 + MgCos20 = 30s² + MgCos20

Also, the force pushing the block up would be 4F/5 = 40s²

So the net work done on the block by the force F and friction for a small displacement of ds is given as:

dw = [40s² - (30s² + MgCos20)_k].ds

Therefore, the work done on the block would be integration of the above expression for S = 0 to S = 1 m

That is, W = dw = [40s² - (30s² + MgCos20)_k].ds = 40/3 - 30 x 0.1 /3 - 20x9.81x0.1Cos20 = -6.1034 J

Now initial kinetic energy = 0.5 mv² = 0.5 x 20 x 4 = 40 J 33.8966

Also, the gravitational force will do some work on the block as it is being pushed upwards. Let us say that block is pushed by s distance above

hence work done by the gravitational force would be MgsSin20 = 67.104 S

gravitational force would itself exhaust all its energy before it gets up by 1 m along the slope. Hence the block would never reach the distance of S = 1 m

NOTE: This is unusual. The most probable explanation is that the coefficient of the kinetic friction is not 0.1 but is rather much less. You might want to recheck.