Question

The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB.
b. WA = WB.
c. WA = 2 WB.
d. WA = 4 WB.
e. WB = 2 WA

Answer 1

Answer:

$WA=2WB$

Explanation:

The work done by a constant force is given by:

$W=F*d*cos(\theta)$

In this case the movement is completetly horizontal so:

$\theta=0$

This reduces the equation to:

$W=F*d$

Where:

$F=Applied\hspace{3}force\hspace{3}to\hspace{3}the\hspace{3}object$

$d=Distance\hspace{3}traveled\hspace{3}by\hspace{3}the\hspace{3}object$

The distance d is given by:

$d=v*t$

Now, with this in mind, for car A:

$WA=F1*v*t1$

Isolating F1:

$F1=\frac{WA}{v*t1}$

For car B

$WB=F2*v*t2$

Isolating F2:

$F2=\frac{WB}{v*t2}$

According to the problem, we know that:

$F1=F2\\t1=2t2$

Therefore, matching F1 and F2, and replacing t1 by 2t2:

$\frac{WA}{v*2*t2} =\frac{WB}{v*t2}$

Multiplying by t2 and v to both sides:

$\frac{WA}{2} =WB$

Finally multiplying by 2 to both sides:

$WA=2WB$

Answer 2

Answer:

Whatever the other guy said

Explanation:

He's right!