Question

A 25.5 kg block is being pulled by a force F = 241 N up along a rough incline. The coefficient of kinetic friction between the block and the incline is μk = 0.40. The block is pulled through a distance of 27.0 m. Angle is 33 degrees.

(a) Solve for the magnitude of the normal force (in Newtons)?

-Calculate the work (in Joules) done by the normal force. Pay attention to the angle in the work formula and the sign of work.

(b) What is the magnitude of the kinetic frictional force (in Newtons)?

- Calculate the work (in Joules) done by the kinetic frictional force. Pay attention to the angle in the work formula and the sign of work.

(c) Calculate the work (in Joules) done by the pulling force F. Pay attention to the angle in the work formula and the sign of work.

(d) Calculate the work (in Joules) done by the gravitational force. Pay attention to the angle in the work formula and the sign of work.

(e) Calculate the total work or net work (in Joules).

(f) Apply the Work-Kinetic Energy Theorem: The block starts from rest. What is its speed (in m/s)after being pulled through 27.0 m?

(Thank you for the help and I will Upvote!!)

Answer 1

a)

from the force diagram , normal force is given as

N = mg Cos33

N = (25.5 x 9.8) Cos33

N = 209.6 N

d = displacement

Work done by the normal force is given as

W_n = N d Cos90                    (Since angle between normal force and displacement is 90)

W_n = N d (0)      

W_n = 0 J

b)

Kinetic frictional force is given as

f_k = mg

f_k = (0.40) (25.5 x 9.8)

f_k = 99.96 N

work done by the kinetic frictional force is given as

W_k = f_k d Cos180

W_k = (99.96) (27) Cos180

W_k = - 2698.92 J

c)

Work done by the pulling force is given as

W_p = F_p d Cos0 = (241) (27) Cos0 = 6507 J

d)

Work done by the gravitational force is given as

W_g = (mg) d Cos(90 - 33) Cos180 = - (25.5 x 9.8) (27) Cos(90 - 33) = - 3674.84 J

e)

Net work done is given as

W = W_n + W_k + W_p + W_g

W = 0 + (- 2698.92) + 6507 + ( - 3674.84)

W = 133.24 J