Question

an object of mass 10kg rests on an inclined plane of 30 degree angle with horizontal. a force P=200 N acts on it upward along the inclined plane and moves it 20m. the coefficient of friction UK=0.25.

2 4 x (cm) 6. An object of mass 10 kg rests on an inclined plane of 30 angle with horizontal. A force 0° angle with horizontal. A force P 200 N acts on it upward along the inclined plane and moves it 20 m. The coefficient of friction = 0.25 Find the work done by: (a) Normal force (b) External force (P) (c) Weight (d) Friction.

Answer 1

a)

F_n = normal force

d = 20 m

$\theta$ = angle between F_n and d = 90

work done by Fn is given as

W = F_n d Cos90

since Cos90 =0

W =0

b)

P = external force = 200 N

d = 20 m

$\theta$ = angle between F_n and d = 0

work done by Fn is given as

W = P d Cos0

W = 200 x 20 Cos0

W = 4000 J

c)

weight = mg = 10 x 9.8 = 98 N

d = 20 m

$\theta$ = angle between weight and "d" = 120

Work done is given as

W = 98 x 20 Cos120

W = -980 J

d)

friction force is given as

f_k = u_k F_n                         where F_n = mg Cos $\theta$

so f_k = u_k mg Cos $\theta$

f_k = (0.25) (10) (9.8) Cos30

f_k = 21.22 N

d = 20 m

$\theta$ = angle between f_k and d = 180    since f_k and d points in opposite direction

so work done is given as

W = f_k d Cos $\theta$

W = 21.22 x 20 Cos180

W = - 424.4 J