Work done on incline with friction

[tehstone]

Homework Statement

A father pushes horizontally on his daughter's sled to move it up a snowy incline, as illustrated in the figure, with h = 4.4 m and θ = 10°. The total mass of the sled and the girl is 35 kg and the coefficient of kinetic friction between the sled runners and the snow is 0.20. If the sled moves up the hill with a constant velocity, how much work is done by the father in moving it from the bottom to the top of the hill?

Homework Equations

W = Fp * d

The Attempt at a Solution

2 things have to be overcome, gravity and the friction.

1. For gravity: F = m*g*h = (35kg)(9.8m/s^2)(4.4m) = 1509 N

2. Friction: W = fk * d.

fk = uk*m*g*cos(10) = (.2)(35kg)(9.8m/s^2)*cos(10) = 67.56

d = h/sin(theta) = 4.4/sin(10) = 25.34

w = 67.56 * 25.34 = 1712 N


So the total force exerted by the father should be 1509N + 1712N = 3221 N but this is wrong. What am I missing?

 

[PhanthomJay]

You are confusing work and force units. The problem asks for the work done by dad, not the force he exerts. What is the correct unit for work??

 

[tehstone]

That would be Joules. and so to go from Newtons to Joules you have to include the distance over which the Newtons were exerted. I did that for the friction calculation. But the work done vs gravity shouldn't need it because gravity is conservative right?

Edit:
So to be clear, for the friction calculation, it is 67.56 N * 25.3 m = 1712 J. But isn't the gravity calculation correct as well? the distance in this case is vertical so it's the height (4.4m) which is included. So then the 1509 N should be Joules. But the numerical answer is the same. This is an online assignment and the units are given, only the magnitude is required, and the figure of 3221 was incorrect.

 

[PhanthomJay]

The work done by gravity is -1509 Joules. Ot if you like, the potential energy change is 1509 Joules. You mistakenly called it a force of 1509 N. Your numerical answer looks correct for the work (not the force) done by dad. Your units for work are not so good.

 

[PhanthomJay]

Oh units were given, sorry. Try using 3200 J, could be a sig fig thing.

 

[tehstone]

I ran out of answer attempts so now I can't find out what the correct answer is. I tried both 3221 J and 3.2e3 J and neither were correct. Oh well, thank you very much for your help.

 

[haruspex]

You overlooked that F was horizontal. That increases the normal force and hence the friction. I believe the correct answer is mgh(t+k_d)/(t(1-t*k_d)) where t = tan(θ) and k_d is the dynamic coefficient of friction. That gives 3342J.
Note that if we write k_d = tan(α) then it becomes mgh tan(θ+α)/tan(θ).

 

[PhanthomJay]

You overlooked that F was horizontal. That increases the normal force and hence the friction. I believe the correct answer is mgh(t+k_d)/(t(1-t*k_d)) where t = tan(θ) and k_d is the dynamic coefficient of friction. That gives 3342J.
Note that if we write k_d = tan(α) then it becomes mgh tan(θ+α)/tan(θ).

I overlooked that also. My apologies.

 

[tehstone]

You overlooked that F was horizontal. That increases the normal force and hence the friction. I believe the correct answer is mgh(t+k_d)/(t(1-t*k_d)) where t = tan(θ) and k_d is the dynamic coefficient of friction. That gives 3342J.
Note that if we write k_d = tan(α) then it becomes mgh tan(θ+α)/tan(θ).

Thank you! That was the correct solution.