Calculating Work Done and Distance Traveled on Inclined Surfaces

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The discussion focuses on calculating work done and distance traveled on inclined surfaces through two example problems involving a crate and a sled. For the crate, the work done by a 100N force is calculated using net forces and displacement, resulting in approximately 139.4 J. The sled problem involves determining how far it moves up an incline, with the net force and kinetic energy leading to a distance of 5.3 m. Participants highlight the importance of considering the component of displacement parallel to the force and the correct signs for forces acting in different directions. The conversation emphasizes the need for clarity in applying physics principles to solve work-related problems effectively.
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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?
 
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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