Calculate work done by friction and by the the weight of a person on an inclined slide?

Question

A person of mass 100kg slides down an inclined slide (60 degrees to the horizontal). The height of the slide is 2.50m and the distance he slides is 5metres. The coefficient of kintic friction(Uk) is 0.1 . Calculate work odne by the person's weight, work done by the normal reaction force from the slide and work done by friction.

Answer 1

Calculate work odne by the person's weight,

WE = FE + PE; work energy is the friction energy plus the lifting energy.  FE = W cos(theta) kS; where W is the person's weight W = mg, theta = 60 deg, k = .1, and S = 5 m.  PE = Wh = mgh; where h = 2.5 m height.  Thus the total work done by the weight (W) of the person is WE = mgkS cos(theta) + mgh = mg(kS cos(theta) + h).  [Note...weight is the driving force for both the friction work and for the decrease in height as the person slides down the ramp.]

work done by the normal reaction force from the slide and

N = W cos(theta) = mg cos(theta) and Sn = 0, which is the distance pushed by the normal weight.  Thus WE = NSn = mg cos(theta)*Sn = 0.  So the normal force does no work as it does not move the person in the normal direction at all.

work done by friction.

FE = NkS = mg cos(theta) kS, as indicated earlier.  [NB:  The work is not done by the friction; friction does not move things.  The work is done by the force of gravity (weight) to overcome friction and move the person along the ramp.]

The physics is this...the total work WE is the sum of the work done by gravity against the friction and done by gravity in lowering the height of the person. to ground level.  It is not, as suggested by another answer, just the vertical work.

Answer 2

49 N = work done by friction

W = F * x

= (Uk * N) * x

= (Uk * m * g * cos(theta)) * x

= (.1 * 100 * 10 * cos(60)) * 5

= (.1 * 100 * 10 * 1/2) * 5

=250 N

using g = 10 m/s^2, although you could use 9.81 if you want.

As for the work done by the weight, you must mean the work done by the force of gravity. Gravity points down, so the distance done in the direction of the force is 2.5 in this case. (Remember that the formula for work is Force (DOT) Displacement, if you are familiar with vectors)

W = F * x

= (m * g) * x

= (100 * 10) * 2.5

= 2500 N

Now for the normal force. It is PERPENDICULAR to the motion, thus the dot product is zero. In other words, the normal force is perpendicular to the surface on which the object rests, so there is no component of the normal force for the direction in which the object moves. Thus no work is zone, 0 N.

Hope that helped :)