How to find the work on an object on an inclined plane pulled with a rope

AI Thread Summary
The discussion focuses on calculating the work done on an object on an inclined plane being pulled by a rope at an angle. Two equations for work are presented: one for an inclined plane and another for an object pulled at an angle, incorporating forces and friction. The calculations yield two different results for work done: 139.8 Joules and 69.9 Joules, but the accuracy of these results is questioned. Participants emphasize the need for clearer explanations of the calculations involved. The overall consensus is that the calculations and their rationale need to be better articulated for clarity.
Kelvin Andersen
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Homework Statement


xllB5.png

g = ##10m/s^2##
##\sqrt{3}## = 1.7

Determine work by the 20N force

Homework Equations


for an inclined plane:

W = ((F+mgsin30)-(mgcos30)μ)s -> (Fx-ff)*s

for object pulled with angle:

W= (Fcos60-(mg-Fsin60)μ)s -> (Fx-ff)*s

The Attempt at a Solution


W= ((Fcos60+mgsin30)-(mgcos30-Fsin60)μ)s
(W= (Fx-ff)*s, Fx=Fcos50+wsin30, ff=(wcos30-Fsin60)*μ)

What I did to get the equation above:
for Fx:
Based on inclined plane equation, Fx=F+wsin30
However, since F is angled, F becomes Fcos60
so Fx=Fcos60+wsin30

for ff:
Based on angled force equation, ff=w-sin60
I change w -> wcos30
so ff=wcos30-Fsin60

sin30=3/s -> s=6
W= (((20)(cos60)+(3)(10)(sin30))-((3)(10)(cos30)-(20)(sin60))(0.2))(6) (inputting number based on the above)

by using √3 = 1.7

W= ((10+15)-(25.5-17)(0.2))(6)

W= (25-1.7)(6) = 139.8 Joules (don't know if correct or not)
 

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Kelvin Andersen said:

Homework Statement


View attachment 215645
g = ##10m/s^2##
##\sqrt{3}## = 1.7

Determine work

Homework Equations


for an inclined plane:

W = ((F+mgsin30)-(mgcos30)μ)s

for object pulled with angle:

W= (Fcos60-(mg-Fsin60)μ)s

The Attempt at a Solution


W= ((Fcos60+mgsin30)-(mgcos30-Fsin60)μ)s

W= (((20)(cos60)+(3)(10)(sin30))-((3)(10)(cos30)-(20)(sin60))(0.2))(3)

by using √3 = 1.7

W= ((10+15)-(25.5-17)(0.2))(3)

W= (25-1.7)(3) = 69.9 Joules (dont' know if correct or not)
Work done by which force you haven't mentioned
 
Kelvin Andersen said:

Homework Statement


View attachment 215645
g = ##10m/s^2##
##\sqrt{3}## = 1.7

Determine work by the 20N force

Homework Equations


for an inclined plane:

W = ((F+mgsin30)-(mgcos30)μ)s

for object pulled with angle:

W= (Fcos60-(mg-Fsin60)μ)s

The Attempt at a Solution


W= ((Fcos60+mgsin30)-(mgcos30-Fsin60)μ)s

W= (((20)(cos60)+(3)(10)(sin30))-((3)(10)(cos30)-(20)(sin60))(0.2))(3)

by using √3 = 1.7

W= ((10+15)-(25.5-17)(0.2))(3)

W= (25-1.7)(3) = 69.9 Joules (don't know if correct or not)

The big problem with what you have posted is that you haven't explained at all what calculations you are carrying out. The line I've underlined is almost impossible to parse.

In any case, your solution looks wrong to me.

Try to explain what you are calculating and why.
 
PeroK said:
The big problem with what you have posted is that you haven't explained at all what calculations you are carrying out. The line I've underlined is almost impossible to parse.

In any case, your solution looks wrong to me.

Try to explain what you are calculating and why.
edited
 
Kelvin Andersen said:
edited

It's still incomprehensible to me.

How, in general, do you calculate the work done by a force?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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