Question

a boy pulls his 6.35kg sled 5m across a patch of ice where the sled accelerates at a=0.39m/s^2. He pulls the sled by a rope with tension 10.6N at an angle of 30 degrees above the horizontal. (a) Find the work done by each force. (b) Find the net work

Answer 1

a)

T = tension force in the rope = 10.6 N

f = frictional force = ?

a = acceleration of sled = 0.39 m/s²

m = mass of sled = 6.35 kg

Along the horizontal direction, force equation is given as

T Cos30 - f = ma

10.6 Cos30 - f = (6.35) (0.39)

f = 6.7 N

For tension force :

W_t = work done by tension force

d = displacement = 5 m

= angle between tension force and displacement = 30

work done by tension force is given as

W_t = T d Cos

W_t = (10.6) (5) Cos30

W_t = 45.89 J

For frictional force :

W_f = work done by frictional force

d = displacement = 5 m

= angle between frictional force and displacement = 180

work done by tension force is given as

W_f = f d Cos

W_f = (6.7) (5) Cos180

W_f = - 33.5 J

For normal force :

W_N = work done by normal force

d = displacement = 5 m

= angle between normal force and displacement = 90

work done by normal force is given as

W_N = N d Cos

W_N = N d Cos90

W_N = 0 J

For force of gravity :

W_g = work done by force of gravity

d = displacement = 5 m

= angle between force of gravity and displacement = 90

work done by force of gravity is given as

W_g = F_g d Cos

W_g = F_g d Cos90

W_g = 0 J

b)

W = net work done

Net work done is given as

W = W_t + W_f + W_N + W_g

W = 45.89 + (- 33.5) + 0 + 0

W = 12.4 J