How much work is done by the gravitational force on the crate?

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SUMMARY

The discussion focuses on calculating the work done by gravitational force on a crate being pulled up an incline. The crate has a mass of 10.6 kg, is pulled with a force of 93.0 N at an angle of 20.1°, and experiences a coefficient of kinetic friction of 0.400. The work done by gravity is calculated as -171 J, while the work done by friction is 198 J, and the work done by the pulling force is 472 J. The final speed of the crate after being pulled 5.08 m is determined to be 2.10 m/s.

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Homework Statement


A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.51 m/s. The pulling force is 93.0 N parallel to the incline, which makes an angle of 20.1° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.08 m.

a) How much work is done by the gravitational force on the crate?
b) Determine the increase in internal energy of the crate-incline system due friction.
c) How much work is done by the 93.0 N force on the crate?
d) What is the change in kinetic energy of the crate?
e) What is the speed of the crate after being pulled 5.08 m?

Homework Equations


Wgravity=-Δ(Ugravity)
Ugravity=m*g*h
W=F*Δr

The Attempt at a Solution


Please excuse lack of units.

a) How much work is done by the gravitational force on the crate?
Since the object is on an incline, the acceleration due to gravity must be adjusted:
cos(20.1) * 9.81 = 9.21

The height of displacement is also needed:
sin(20.1) = x/5.08
x=1.75

Wgravity=-Δ(Ugravity)
Wgravity=-(mghf-mghi)
Wgravity=-(10.6*9.21*1.75 - 10.6*9.21*0)
Wgravity= -171J

I know that this answer is incorrect but do not know why.

b) Determine the increase in internal energy of the crate-incline system due friction.
Δ(KE) = Wfriction
Wfriction= Fk * Δr
Fkk * N
Fk=0.400 * (9.21)(10.6)
Fk=39.0
Wfriction= 39 * 5.08
W=198J=Δ(KE)

c) How much work is done by the 93.0 N force on the crate?
Wf=F*Δr
Wf=93 * 5.08
Wf=472J

d) What is the change in kinetic energy of the crate?
ΣW = Δ(KE) + Δ(Ug)
Wfriction + Wf = Δ(KE) + Δ(Ug)
-198J + 472 = Δ(KE) + 171J
Δ(KE) = 103J

e)What is the speed of the crate after being pulled 5.08 m?
Δ(KE) = (1/2)*m*vf^2 - (1/2)*m*vi^2
103 = (1/2) * 10.6 * vf^2
Vf=2.10m/s
 
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Wgravity=-(10.6*9.21*1.75 - 10.6*9.21*0)
Wgravity= -171J


Welcome to PF.

The acceleration due to gravity is still 9.8 m/s². It is potential energy. It is a conservative field and gravity acts at all points in the downward direction against which it has moved to that height.

Correct for the use of 9.21 elsewhere.
 

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