# Homework Help: Work questions- An object on an inclined plane involving friction.

1. Apr 2, 2014

### Dtbennett

1. The problem statement, all variables and given/known data

A skier, mass m = 85 kg including skis, poles and clothes, skis directly down a slope with an angle Φ = 21° to the horizontal. His body is rigid, the poles are not touching the snow. He is travelling at a constant speed of 71 kph in a straight line. The coefficient of kinetic friction is 0.21. In all of the following, think about the units and sign, and it probably helps to draw a free body diagram.

a)What is the rate at which gravity is doing work on the skier? (Here, 'skier; includes skis, poles and clothes.) (in kW)
b)What is the rate at which friction is doing work on him?
c)What is the rate at which air resistance is doing work on him?

2. Relevant equations

Work done by gravity= mgdcos(x)
kW= J/s
F= ma

3. The attempt at a solution

I got the work equation from a textbook, but I think the problem is rate. Subbing in the values gives
85x 9.8m/s^2 x 19.7222 m/s x cos(11)
= 1624.6111 kg/s
Converting this to kW, you get 161.109.
But this isn't really a rate.... How do you figure out the rate? is it a derivative of some sort?

2. Apr 2, 2014

### HallsofIvy

First, are you sure about "Work done by gravity = mgdcos(x)"? Gravity works vertically and if x is the angle of the slope, 21 degrees here, cos(x) will give a horizontal factor not vertical.

Second, since the problem does say that the slope is 21 degrees, why are you using cos(11)?

Third, to find the rate at which work is done, dW/dt, use the "chain rule"- dW/dt= (dW/dx)(dx/dt) where "x" is the distance moved down the slope (what you called, I think, "d" in your formua). You are told that "He is travelling at a constant speed of 71 kph" so dx/dt= 71.

3. Apr 3, 2014

### BvU

Hello dt and welcome to PF.
So rate of something is just amount of something per unit time. The giveaway in the problem is that they want it expressed in kW which is kJ/s.
Your expression (with sin(21) instead of cos(11) ) is right, but you should take better care of the dimensions:

kg * m/s2 * m/s = N * m / s = J/s = W and not kg/s. And converting W to kW is not a matter of dividing by 10.

Now do the same exercise for the friction.

Show the free body diagram and try to find the wind rate of work in two "different" ways so that you have a check on the result.

Last edited: Apr 10, 2016
4. Apr 3, 2014

### dauto

Also kph isn't an accepted abbreviation. km/h is the only accepted abbreviation for kilometer per hour.

5. Apr 9, 2016

### blueray101

Hello, I am actually doing this question as well. I am up to this bit:

b)What is the rate at which friction is doing work on him?

I am a bit confused about the direction of the velocity for friction. I assume its velocity is parallel to the incline?

This is what I have for part b so far. I'll use the values the OP has posted.
u=kinetic friction
u*N=Ff
Ff=0.21*85*cos(21)*-9.8

Power=P=fv=Ff*v

v=71/3.6 m/s

=0.21*85*cos(21)*-9.8*71/3.6

Now, this is where I am unsure. I have assumed that the frictional velocity is the one parallel to the incline that is 71km/h. Is this correct?

6. Apr 10, 2016

Yes.