Question

8. A physics TA pushes a 20.0 kg cart at a constant velocity 22.0 m down the hall. The force she exerts on the cart is directed at an angle 26.0° below the horizontal, and the force of friction opposing the motion of the cart is 44.0 N. What is the total work done on the cart?

Answer 1

Answer:

Total work: 0 J

Explanation:

First of all, we have to find the magnitude of the force that is pushing the cart forward.

We can do that by writing the equations of motion along the horizontal direction; we have:

$F cos \theta - F_f = ma$

where

$F cos \theta$ is the horizontal component of the force of push, where

F is the magnitude of the force

$\theta=-26^{\circ}$ is the angle of the force with the horizontal

$F_f=44.0 N$ is the force of friction

m = 20.0 kg is the mass of the cart

a = 0 is the acceleration, since the cart is moving with constant velocity

Solving for $F cos \theta$,

$Fcos \theta - F_f = 0\\F cos \theta = F_f = 44.0 N$

This component of the pushing force is the one doing work on the cart, so the work done on the cart by the force F is

$W_F = Fcos \theta d$

where

d = 22.0 m is the displacement of the cart

Substituting,

$W_F=(44.0)(22.0)=+968 J$

However, there is another force acting in the horizontal direction: the force of friction, acting in the direction opposite to the displacement. So, the work done by friction will be negative, and it is:

$W_f=-F_f d =-(44.0)(22.0)=-968 J$

This means that the total work done on the cart is:

$W=W_F+W_f=+968+(-968)=0 J$