Now let\'s consider the total work done on an object that has several forces act

Question

Now let's consider the total work done on an object that has several forces acting on it. A tractor is hitched to a sled loaded with firewood and pulls the sled a distance of 20.0 m along level frozen ground (Figure 1) . The total weight of the sled and load is 14,700 N. The tractor exerts a constant force F T with magnitude 5000 N at an angle of = 36.9 above the horizontal, as shown. A constant 3500 N friction force opposes the motion. Find the work done on the sled by each foce individually and the total work done on the sled by all the forces.

SOLUTION

SET UP (Figure 2) shows a free-body diagram and a coordinate system, identifying all the forces acting on the sled. As in the preceding example, we point the x axis in the direction of displacement.

SOLVE The work Ww done by the weight is zero because its direction is perpendicular to the displacement. (The angle between the two directions is 90, and the cosine of the angle is zero.) For the same reason, the work Wn done by the normal force n  (which, incidentally, is not equal in magnitude to the weight) is also zero. So Ww=Wn=0 .

That leaves FT and f. From W=Fs, the work WT done by the tractor is (with F=FTcos)

WT==(FTcos)s(5000N)(0.800)(20.0m)=80,000Nm=80.0kJ

The friction force f is opposite to the displacement, so, for this force, = 180 and cos = 1. The work Wf done by the friction force is

Wf==(fcos180)s=(3500N)(20.0m)70,000Nm=70.0kJ

The total work Wtotal done by all of the forces on the sled is the algebraic sum (not the vector sum) of the work done by the individual forces:

Wtotal==WT+Ww+Wn+Wf80.0kJ+0+0+(70.0kJ)=10.0kJ

Alternative Solution: In the alternative approach, we first find the vector sum (resultant) of the forces and then use it to compute the total work. The vector sum is best found by using components.

From (Figure 2) ,

Fx=(5000N)cos36.93500N=500N

Fy=(5000N)sin36.9+n+(14,700N)

We don't really need the second equation; we know that the y component of force is perpendicular to the displacement, so it does no work. Besides, there is no y component of acceleration, so Fy has to be zero anyway. The work done by the total x component is therefore the total work:

Wtotal=(500N)(20.0m)=10,000J=10.0kJ

This is the same result that we found by computing the work of each force separately.

REFLECT Note that the free-body diagram of the sled being pulled by the tractor is exactly the same as the free-body diagram we would draw for a box being pulled to the right across a rough horizontal table by a rope. If we employ the same thinking we used to determine the sign of the work done by each of the forces acting on the box, we can check whether the signs of our calculations for the work done by each force on the sled make sense. WT should be positive (which it is), Wf should be negative (which it is), and Ww and Wn should be zero. A natural question to ask is: Where did the energy associated with Wtotal go? We will return to this question later in this chapter.

Part A - Practice Problem:

Suppose the tractor pulls horizontally on the sled instead of at an angle of 36.9. As a result, the magnitude of the friction force increases to 4900 N . What is the total work done on the sled?

Express the work in joules to three significant figures.

Explanation / Answer

Work done by friction = f.s = 3500*20*(-1) = -70,000 J
Work done by Force = F.s = 5000*20*cos(36.9) = 79,968.465 J
Net Work = 9968.465 J
A. Work done by froiction = f.s = -4900*20 = -98000 J
Work done by force = F.s = 5000*20 = 100000 J
Net Work = 2000 J

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