can anybody find work done?

Question

when a body of mass m moves up on an incline plane of angle A through a distance s show that the work done against the friction is umgcosA.s where u is the coeficient of friction of plane

Answer 1

The work function is defined to be W = Fs; where F = uN = umg cos(A) using your symbols. F is the force applied parallel to the inclined plane as the block is shoved upward along the plane. The uN is the frcitional force that F has to overcome while pushing the block up the plane.

Weight is just mg, the block mass times the acceleration due to gravity. N is the so-called normal weight of the block. A normal weight is that vector of weight pressing perpendicular into the plane; as the plane is inclined A degrees, the perpendicular (normal) part of the block's weight N = mg cos(A). [NB: You can test this by noting that when there is no incline and A = 0 degrees, we have N = mg showing the normal weight and the total weight of the block are the same value.]

This is really a fundamental question testing to see if you understand the work function, W = Fs. The important thing to remember is that force is a vector and, in this case, we are interested in the force pointed parallel to and upward on the inclined plane, and equal to the counteracting frictional force.

By the by, the work done against friction is not the only work done here. There is also work against gravity as the block gets higher off the ground as it moves up the plane.

Answer 2

To find the work done by friction, we must know first the frictional force acting on it.

Frictional force is the some multiple or submultiples of the normal reaction acting on the object.

If N is the normal reaction then frictional force f = μ N.
μ is the multiplication factor or the coefficient of friction.

The normal reaction is the reaction by the plane exerted on the object by the plane.

The weight of the object mg is acting vertically down.

That is to say that the weight is pressing the plane vertically down, its component acting perpendicular to the plane is mg cos θ, where θ is the inclination of the plane.

{Note that when the plane is turned through an angle θ from its horizontal position, the perpendicular to the plane tilts through an angle θ}

Since mg cos θ is the force with which the object presses the plane, and also since the object is not moving along this direction, the normal reaction N = mg cos θ

Therefore, the frictional force acting on the object is

f = μ N. = μ mg cos θ

If the object moves through a distance S, the work done by a force is the product of the force and the displacement in the direction of the force.

Since friction is always against the direction of displacement the work done is

─ μ mg cos θ * S.
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Answer 3

its complex a bit, but i can explain to you when u contact me soonest possible.