A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
of 20 m along level ground (Figure 3). The total weight of sled and load is 14,700
2
N. The tractor exerts a constant 5000 N force at an of 36.9
◦ angle of above the
horizontal. A 3500 N friction force opposes the sled’s motion. Find the work
done by each force acting on the sled and the total work done by all the forces.

Respuesta :

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

Work done by the applied force

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

Work done by frictional force

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

Net work done by all the forces on the tractor

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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