Calculating Work Done and Distance Traveled on Inclined Surfaces

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Homework Help Overview

The discussion revolves around calculating work done and distance traveled on inclined surfaces, specifically involving forces acting on a crate and a sled on inclines. The subject area includes concepts from mechanics, particularly work, energy, and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done by various forces, including gravitational force and applied force. Questions arise regarding the application of the dot product and the correct interpretation of forces acting on the objects. There is also a focus on the signs of forces and displacements in the context of work calculations.

Discussion Status

Some participants have offered guidance on the need to consider the component of displacement parallel to the forces involved. Others have raised questions about the completeness of the problems and whether additional parts exist. The discussion reflects a mix of attempts to clarify concepts and explore different aspects of the problems presented.

Contextual Notes

Participants note potential confusion regarding the necessity of certain information for the calculations and the implications of sign conventions in the context of work and energy. There is an acknowledgment that the original poster may not have studied certain mathematical concepts relevant to the problems.

willingtolearn
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#1
A 10kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100Nparallel to the incline, which makes an angle of 20* with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m.
a)HOw much work is done by the 100N
----------------
Fg = 98 N
Fn = 92.1 N
Fpar = 33.5 N
Ffricton = 36.84 N
Fa = 100 N
Fnet = 100 - 33.5 - 36.84 = 29.66 N
W=F.d
d (in the vertical) = 4.7 m
a) W=29.66 (4.7) 139.402 J

#2
In a circus performance, a monkey on a sled is given an initial speed of 4 m/s up a 20* incline. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between sled and incline is .20. How far up the incline does the sled move ?
--------------
Fn = 184.2 N
Fpar = 67.03 N
Ff = 36.84 N
Fnet = Fpar - Ff = 30.19
KE = .5(20)(4)^2 = 160 J
KE = W
W = F.d
160 = 30.19 d
d = 5.3 m

What i did wrong on those 2 problem ?
 
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For #1

Have you studied dot product?

The work done by the 100N force is 100N times the component of displacement parallel to the 100N force... what is the component of displacement parallel to the 100N force?

For #2

Fpar and Ff are acting in the same direction. be consistent with your signs... down the plane negative... up the plane positive...

Net work done = change in kinetic energy

Net work done = -160
 
I'm wondering, are there more parts to problem #1? Because there's a lot of information that's not necessary for part a).
 
#1 No, I haven't studied the dot product yet
A) HOw much work is done by the gravitational force ?
b) Ho much work is done by 100 N force ?
c) What is the change in kinetic energy of the crate ?
d) What is the speed of the crate after it is pulled 5 m?

Sorry about that
 
Last edited:
willingtolearn said:
#1 No, I haven't studied the dot product yet
A) HOw much work is done by the gravitational force ?
b) Ho much work is done by 100 N force ?
c) What is the change in kinetic energy of the crate ?
d) What is the speed of the crate after it is pulled 5 m?

Sorry about that

Ah... I see. Ok... for all these work problems... to get the work done by a force... you need the force * (component of displacement parallel to the force)

starting with a)... you need:

the gravitational force * displacement parallel to the gravitational force

so you need displacement in the up/down direction (since that is the direction of the force)... take signs into account for both the force and displacement... take up positive and down negative...

so what is the work done by the gravitational force?
 

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