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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

W

_{gravity}=-Δ(U

_{gravity})

U

_{gravity}=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

W

_{gravity}=-Δ(U

_{gravity})

W

_{gravity}=-(mgh

_{f}-mgh

_{i})

W

_{gravity}=-(10.6*9.21*1.75 - 10.6*9.21*0)

W

_{gravity}= -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) = W

_{friction}

W

_{friction}= F

_{k}* Δr

F

_{k}=μ

_{k}* N

F

_{k}=0.400 * (9.21)(10.6)

F

_{k}=39.0

W

_{friction}= 39 * 5.08

W=198J=Δ(KE)

c) How much work is done by the 93.0 N force on the crate?

W

_{f}=F*Δr

W

_{f}=93 * 5.08

W

_{f}=472J

d) What is the change in kinetic energy of the crate?

ΣW = Δ(KE) + Δ(U

_{g})

W

_{friction}+ W

_{f}= Δ(KE) + Δ(U

_{g})

-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*v

_{f}^2 - (1/2)*m*v

_{i}^2

103 = (1/2) * 10.6 * v

_{f}^2

V

_{f}=2.10m/s