Calculating work done - Need help with this problem

[ncondon]

Homework Statement

A 12kg box is pulled 7m along a 35 degree incline by a person pulling a rope that makes a 20 degree angle with the incline. The tension in the rope is 100N and the coefficient of kinetic friction between the box and incline is 0.3

Calculate the work done by:
(i) Tension in rope
(ii) Gravity
(iii) Friction
(iv) normal force

Homework Equations

Wd = Fcos(angle in degrees) * distance

The Attempt at a Solution

Wd Tension = Tcos(angle) * distance - I'm not sure which angle to use for this one
= 100cos(20) * 7m
Wd gravity = -mgcos(angle) * distance
= - 12kg * 9.8 *cos(35) * 7
wd Friction = coeffiicientKE *mg* cos(angle) * distance
= 0.3 * 12 * 9.8 cos(35) * 7
wd Normal Force = mg cos(angle) * distance
= 12 * 9.8 cos 35 * 7

Is this right. Can anyone verify that it is or make the appropriate corrections please

 

[dx]

First one is correct, the rest are wrong.

The angle θ in the formula W = F⋅cos(θ)⋅d is the angle between the force and the displacement. For example, for the second one, if you draw a picture it's pretty easy to see that the angle between the gravity force and the displacement is 90 + 35, so W = F⋅cos (90 + 35)⋅d = mg⋅(-sin(35))⋅d.

 

[ncondon]

ok thanks for that.

For the WD due to normal force i have worked out as it is negative to the work done due to gravity. Is that correct?

I'm not too sure how to do work done due to friction. I'm pretty sure it's not simply WDfriction = uk * mg * distance or would it be WDfriction = uk * mg* sin 45.

 

[dx]

For the WD due to normal force i have worked out as it is negative to the work done due to gravity. Is that correct?

No. The normal force is perpendicular to the displacement, so θ = 90 and cos(θ) = 0.

I'm not too sure how to do work done due to friction. I'm pretty sure it's not simply WDfriction = uk * mg * distance or would it be WDfriction = uk * mg* sin 45.

What is the direction of the friction force? Why did you choose θ to be 45?

 

[ncondon]

so the work done due to normal foce is 0 then since cos(90) =0?

come to thnk of it the work done due to friction should have been WDfriction = uk * mg * cos(0) therefore it should just simply be WDfriction = uk * mg?

 

[dx]

so the work done due to normal foce is 0 then since cos(90) =0?

Yes.

come to thnk of it the work done due to friction should have been WDfriction = uk * mg * cos(0) therefore it should just simply be WDfriction = uk * mg?

No. What is the magnitude of the normal force? (It's not mg.)

 

[ncondon]

well it wouldn't be the 100N from the tension force. other than that I'm not too sure what the normal force would be since i thought it would be mg. since I can't seem to find other information in the question. would the angle have been 0 but

 

[ncondon]

or is is -mg?

 

[dx]

The normal force is the normal (perpendicular) force exerted by the inclined plane on the block, and is perpendicular to the inclined plane. Find the total force on the block due to gravity and tension, and find it's component perpendicular to the inclined plane. Since the block stays on the inclined plane, the normal force must balance this component, and this condition determines the normal force.

 

[ncondon]

so to balance the forces out we do the equation:

Fgsin(-35) + Tcos(20) + n = 0

 

[dx]

No, how did you get that? The component of the tension force perpendicular to the incline is Tsin(20), and the component of the weight perpendicular to the incline is -mgcos(35).

 

[ncondon]

well i have worked n = mgcos(35) - Tsin(20). so is the frictional force * -1.

 

[ncondon]

oops i meant to say *0.3 then the distance

 

[dx]

Frictional force is normal force times the coefficient of kinetic friction, pointing opposite to the direction of motion.

 

[ncondon]

Frictonal force = 0.3(mgcos(35) - Tsin(20)) * -1

WdFriction = Frictional force * 7m

Is that right

 

[dx]

Well, the magnitude of the frictional force is just 0.3(mgcos(35) - Tsin(20)) without with -1, but you have to multiply by -1 when you calculate the work because it is pointing opposite to the displacement (cos(180) = -1).

 

[ncondon]

so eventually i have come up with the right calculations

 

[dx]

Yes.