Starting from rest, a 5.4kg block slides...? MOST OF WORK DONE?

Question

Starting from rest, a 5.4 kg block slides 2.8 m down a rough 30.0° incline. The coefficient of kinetic friction between the block and the incline is µk = 0.436.

*the work done by the force of gravity is 74.1636J
*the work done by normal force is 0J

Can you help me determine the work done by the friction force between block and incline?

Answer 1

The force of friction is F = Nk; where N is the normal weight = mg cos(theta) and k = .436, the coefficient of friction. m is the mass, g the acceleration due to gravity (9.81 m/sec^2 on Earth's surface), and theta = 30 deg, the ramp's incline angle.

Work = Fd = kmg cos(theta) d; where d is the distance = 2.8 m, which I assume you meant that distance along the ramp's surface. You have all the numbers for the factors in the work equation; so you can do the math.

Lesson, all those other work values do not fit into the work done against the friction. Also, "normal force" is really N, the normal weight. And since it points directly into the ramp and the block does not move in that direction, one would expect there to be no work done by the normal weight of the block.

The force of gravity in fact is the weight of the block W = mg. And, while that pulls at the block along P = W sin(theta) so that f = P - F, it does not figure into the work done by friction. Total net work, however, would be fd = (P - F)d, but that's not what you asked for.

If you want total net work, you can figure that out (you have all the numbers) for doing fd = (P - F)d ~ (74 J - kmg cos(theta) d).

I am curious what kind of textbook would refer to the weight of a block as the force of gravity. It is, of course, but most people would better recognize it as weight.

Answer 2

But for friction the gravity will do a work of

Mg L sin 30 = 5.4 x 9.8 x 2.8 x 0.5 = 74.088J

The excess work (74.1636 ─74.088 ) is done by frictional force = 0.0756 J