Question

Figure 3 shows a 40 kg crate that is being pushed with constant velocity at a distance 8.0 m along a 30° incline plane by the horizontal force of 500N. The coefficient of kinetic friction between the crate and the incline, µk is 0.42. Assuming the crate is moving upwards towards the right, calculate the work done by the applied force, the frictional force and the gravitational force and the net work done on the crate.

Answer 1

Answer: Wfr = -1883.25 J

Step-by-step explanation:

1) In the x direction (direction of motion of the crate):

Fcos(30^o) - f_fr - mgsin(30^o ) = 0 (because a_x = 0)

f_fr = Fcos(30^o ) - mgsin(30^o)

f_fr = (498)(√3/2) - (40)(9.81) (1/2) = 235.28 N

⇒ W_fr = f_fr.s.cos(180^o) = -(235.28)(8.0) = -1882.25 J (1)

2) In the y direction (perpendicular to the motion of the crate):

n - Fsin(30^o) - mgcos(30^o) = 0 (because a_y = 0)

⇒ n = Fsin(30^o) + mgcos(30^o) = (498)(1/2) + (40)(9.81)(√3/2) = 588.83 N

⇒ f_fr = μ_k.n = (0.40)(588.83) = 235.53 N

⇒ W_fr = f_fr.s.cos(180^o ) = -(235.53)(8.0) = -1884.24 J (2)

Average of calculations (1) & (2): W_fr = -1883.25 J

Answer 2

Based on the given figure, you have that the components of the forces on the direction of the incline are:

applied force:

Fa = 500N*cos(-30°) = 433.01N

friction force:

Fr = µk*N = µk*m*g*cos(30°) = (0.42)(40kg)(9.8 m/s^2)cos(30°) = 142.58N

gravitational force:

Fg = m*g*sin(30°) = (40kg)(9.8)sin(30°) = 196.0N

Next, consider that the work done by a force is equal to the product of the force and the distance at which the force is applied. In this case, such distance is 8m.

Then, you have:

Work done by the applied force:

WFa = (433.01N)(8.0m) = 3464.08 J

Work done by the friction force:

WFr = (142.58N)(8.0m) = 1140.64 J

Work done by the gravitational force:

WFg = (196.0N)(8.0m) = 1568.0 J

Finally, the net work, based on the direction of the forces, is:

Net work = WFa - WFr - WFg = 3464.08J - 1140.64J - 1568.0J = 755.44J

Hence, the net work is 755.44 J