A 4 kg block is lowered down a 37 degree incline a distance of 5 m from point A
ID: 2194784 • Letter: A
Question
A 4 kg block is lowered down a 37 degree incline a distance of 5 m from point A to point B. A horizontal force (F=10N) is applied to the block between A and B as shown in the figure. The kinetic energy of the block at A is 10J and at B is 20J. How much work is done on the block by the force of friction between A and B.
A 4 kg block is lowered down a 37 degree incline a distance of 5 m from point A to point B. A horizontal force (F=10N) is applied to the block between A and B as shown in the figure. The kinetic energy of the block at A is 10J and at B is 20J. How much work is done on the block by the force of friction between A and B.Explanation / Answer
Well, the total energy input is the gravity work mgh and the
horizontal work 10N * distance in horizontal PLUS 10J of
initial kinetic energy
The energy in the system at the end is 20J PLUS all the heat
which is the frictional work.
So,
Friction work =10N*distance in x + mgh + 10J - 20J
distance in x = 4 m
h(height change) = 3 m
So, Friction = 10N*4m + 4kg * 9.8 m/s^2*3m - 10 J
I get 147.6 Joules
the great majority of the work goes into friction