WORK DONE BY SEVERAL FORCES
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance of 20 m along level ground (Fig. 6.7a). The total weight of sled and load is 14,700 N. The tractor exerts a constant 5000-N force at an angle of 36.9° 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.
IDENTIFY and SET UP:
Each force is constant and the sled’s displacement is along a straight line, so we can use the ideas of this section to calculate the work. We’ll find the total work in two ways: (1) by adding the work done on the sled by each force and (2) by finding the work done by the net force on the sled. We first draw a free-body diagram showing all of the forces acting on the sled, and we choose a coordinate system (Fig. 6.7b). For each force—weight, normal force, force of the tractor, and friction force—we know the angle between the displacement (in the positive x-direction) and the force. Hence we can use Eq. (6.2) (W=F s \cos \phi) to calculate the work each force does.
As in Chapter 5, we’ll find the net force by adding the components of the four forces. Newton’s second law tells us that because the sled’s motion is purely horizontal, the net force can have only a horizontal component.
EXECUTE:
(1) The work W_w done by the weight is zero because its direction is perpendicular to the displacement (compare Fig. 6.4c).
For the same reason, the work W_n done by the normal force is also zero. (Note that we don’t need to calculate the magnitude n to conclude this.) So W_w = W_n = 0.
That leaves the work W_T done by the force F_T exerted by the tractor and the work W_f done by the friction force f. From Eq. (6.2) (W=F s \cos \phi),
W_{\mathrm{T}}=F_{\mathrm{T}} s \cos 36.9^{\circ}=(5000 \mathrm{~N})(20 \mathrm{~m})(0.800)=80,000 \mathrm{~N} \cdot \mathrm{m}= 80 kJ
The friction force \vec{f} is opposite to the displacement, so for this force Φ = 180° and cos Φ = -1. Again from Eq. (6.2),
W_{f}=f s \cos 180^{\circ}=(3500 \mathrm{~N})(20 \mathrm{~m})(-1)=-70,000 \mathrm{~N} \cdot \mathrm{m} = -70 kJ
The total work W_{tot} done on the sled by all forces is the algebraic sum of the work done by the individual forces:
W_{\mathrm{tot}}=W_{w}+W_{n}+W_{\mathrm{T}}+W_{f}=0+0+80 \mathrm{~kJ}+(-70 \mathrm{~kJ}) = 10 kJ
(2) In the second approach, we first find the vector sum of all the forces (the net force) and then use it to compute the total work. It’s easiest to find the net force by using components. From Fig. 6.7b,
\Sigma F_{x}=F_{\mathrm{T}} \cos \phi+(-f)=(5000 \mathrm{~N}) \cos 36.9^{\circ}-3500 \mathrm{~N}= 500 N
\sum F_{y}=F_{\mathrm{T}} \sin \phi+n+(-w)= (5000 N) sin 36.9° + n – 14,700 N
We don’t 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 \Sigma F_{y} must be zero anyway. The total work is therefore the work done by the total x-component:
W_{\mathrm{tot}}=\left(\sum \overrightarrow{\boldsymbol{F}}\right) \cdot \vec{s}=\left(\sum F_{x}\right) s=(500 \mathrm{~N})(20 \mathrm{~m})=10,000 \mathrm{~J}= 10 kJ
EVALUATE: We get the same result for W_{tot} with either method, as we should. Note that the net force in the x-direction is not zero, and so the sled must accelerate as it moves. In Section 6.2 we’ll return to this example and see how to use the concept of work to explore the sled’s changes of speed.
