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

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Homework Help Overview

The discussion revolves around calculating the work done by a force on an object on an inclined plane, specifically focusing on a 20N force. The context includes the use of gravitational acceleration and trigonometric functions related to angles of inclination.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to derive equations for work based on different forces acting on the object, including friction and gravitational components. There are multiple formulations presented, with some participants questioning the clarity and correctness of the calculations. Others raise concerns about the lack of explanation regarding the calculations being performed.

Discussion Status

The discussion is ongoing, with participants providing various attempts at formulating the work equation. Some have expressed uncertainty about the correctness of their calculations, while others are seeking clarification on the methodology used to arrive at those equations. There is a recognition of the need for clearer explanations of the steps taken in the calculations.

Contextual Notes

Participants note that the problem involves specific angles and forces, and there is mention of using approximations for trigonometric values. The discussion highlights the importance of understanding the underlying principles of work calculation in the context of forces acting on an inclined plane.

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?
 

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