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

[KD-jay]

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

 

[LowlyPion]

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.

 

[Moetasim]

I would like to point out that there are a few errors in the calculations and assumptions made in this solution attempt.

Firstly, the formula used for work done by gravity, Wgravity=-Δ(Ugravity), is incorrect. This formula assumes that the crate is being lifted vertically, but in this case, it is being pulled up an incline. The correct formula for work done by gravity in this scenario would be Wgravity=mgΔh, where Δh is the change in height of the crate. In this case, the crate is being pulled up the incline, so the change in height would be 1.75 m, not 5.08 m.

Secondly, the value for the acceleration due to gravity, 9.81 m/s^2, should not be adjusted for the incline. The acceleration due to gravity remains constant regardless of the angle of the incline.

Thirdly, the calculation for the work done by the 93.0 N force on the crate is incorrect. The formula used, Wf=F*Δr, does not take into account the angle of the force with respect to the displacement. The correct formula would be Wf=F*cosθ*Δr, where θ is the angle between the force and displacement vectors. In this case, the angle is given as 20.1°, so the work done by the force would be 93*cos20.1*5.08 = 440J.

Finally, the calculation for the change in kinetic energy of the crate is also incorrect. The formula used, ΣW = Δ(KE) + Δ(Ug), does not take into account the work done by friction. The correct formula would be ΣW = Δ(KE) + Wfriction + Wf, where Wfriction and Wf are the work done by friction and the force, respectively. In this case, the correct calculation would be 103J = (1/2)*m*vf^2 - (1/2)*m*1.51^2 + 198 + 440, which gives a final speed of 2.44 m/s.

In conclusion, as a scientist, I would suggest double-checking all assumptions and calculations and using the correct equations for the given scenario to ensure accurate and precise results.