How do I solve for work done using applied force and distance on an incline?

[trogdor5]

Homework Statement

Homework Equations

W = fdcos∅

The Attempt at a Solution

I know how to solve the problem using potential energy (i.e. mgh) and the answer is 4.41 J.

My question is, how do I solve it using the work done by the applied force using fdcos∅?

The work done parallel to the incline is 82sin(53) = 65.49
The distance the the box moves up the incline when h = .15 is (I think), .15cos(53) = .09027.

However, when multiplying them together (.09027*65.49) I get 5.9 J.

What am I doing wrong? I'm pretty sure it's in finding the distance that the box travels up the incline.
-edit- I see what I'm doing wrong and it is in the distance the box travels up. I need the angle of the incline, not the force to ground. Is there any way to find that angle?

 

[PhanthomJay]

Homework Statement

Homework Equations

W = fdcos∅

The Attempt at a Solution

I know how to solve the problem using potential energy (i.e. mgh) and the answer is 4.41 J.

My question is, how do I solve it using the work done by the applied force using fdcos∅?

The work done parallel to the incline is 82sin(53) = 65.49
The distance the the box moves up the incline when h = .15 is (I think), .15cos(53) = .09027.

However, when multiplying them together (.09027*65.49) I get 5.9 J.

What am I doing wrong? I'm pretty sure it's in finding the distance that the box travels up the incline.
-edit- I see what I'm doing wrong and it is in the distance the box travels up. I need the angle of the incline, not the force to ground. Is there any way to find that angle?

no, the angle of the incline is not given and cannot be calculated . Does it matter?

 

[trogdor5]

No, it's fine. I was just wondering I guess. Thank you :)

 

[PhanthomJay]

Well wait a second, the comp of the force parallel to the incline must be equal to the comp of the weight parallel to the incline. That is, with some trig, mgsinψ = Fsin(53-ψ). Or, mg/F = sin(53-ψ)/sinψ. Here, ψ is the angle of the incline. So cranking out the numbers, 3(9.8)/82 = 0.36 = sin(53-ψ)/sinψ. i would think you can solve this for ψ , but how? I don't know!

 

[ap123]

I think there's 2 ways of doing this:

(1) The change in KE is 0, so the total work done on the box is zero.
So the work done by F is the negative of the work done by gravity.
W_gravity = -mgh ( h is the vertical height gained by the box ).
So
W_force = mgh = 4.41J

(2) Put x-axis along the ramp pointing up.
Then
F_net_x = Fcosα - wsinθ = 0
( α is angle F makes with the x-axis, and θ is the angle the incline makes with the horizontal surface ).
So, Fcosα = wsinθ
So, W_force = Fdcosα, (where d is the distance the box travels along the incline ).
→ W_force = Fdwsinθ/F = dwsinθ
Using some trig, d = h/sinθ
→W_force = hwsinθ/sinθ = homework = mgh = 4.41J

 

[PhanthomJay]

I think there's 2 ways of doing this:

(1) The change in KE is 0, so the total work done on the box is zero.
So the work done by F is the negative of the work done by gravity.
W_gravity = -mgh ( h is the vertical height gained by the box ).
So
W_force = mgh = 4.41J

(2) Put x-axis along the ramp pointing up.
Then
F_net_x = Fcosα - wsinθ = 0
( α is angle F makes with the x-axis, and θ is the angle the incline makes with the horizontal surface ).
So, Fcosα = wsinθ
So, W_force = Fdcosα, (where d is the distance the box travels along the incline ).
→ W_force = Fdwsinθ/F = dwsinθ
Using some trig, d = h/sinθ
→W_force = hwsinθ/sinθ = homework = mgh = 4.41J

Oh yes, sorry, i was focusing on determining the angle of the incline instead. The angle that the incline makes with the horizontal is of course____?____ degrees. ?

 

[ap123]

As you've already pointed out :
mgsinψ = Fsin(53-ψ)
Expand out the sin(53-ψ), divide through by sinψ, leaving the only unknown as tanψ, leading to ψ = 39.7°