(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.
<h3>
Work done by the applied force</h3>
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
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
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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Answer:
4 m/s² down
Explanation:
We'll begin by calculating the net force acting on the object.
The net force acting on the object from the left and right side is zero because the same force is applied on both sides.
Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:
Force up (Fᵤ) = 15 N
Force down (Fₔ) = 25 N
Net force (Fₙ) =?
Fₙ = Fₔ – Fᵤ
Fₙ = 25 – 15
Fₙ = 10 N down
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (ml= 2.5 Kg
Net force (Fₙ) = 10 N down
Acceleration (a) =?
Fₙ = ma
10 = 2.5 × a
Divide both side by 2.5
a = 10 / 2.5
a = 4 m/s² down
Therefore, the acceleration of the object is 4 m/s² down
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