Question

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A second hand on the right side of the block is pushing it to the left. The block starts from rest. Then Hand 1 pushes with a force of 8 N and the block moves to the right a distance of 4 m, where it has a final velocity of 2 m/s. Between its initial and final positions, how much work did Hand 2 do?

Answer 1

Answer:

$W_2=-12J$

Explanation:

The work of force 2 will be given by the vectorial equation $W_2=F_2.d$. We know the value of $F_1$ and have information about its movement, which relates to the net force $F=F_1+F_2$.

About this movement we can obtain the acceleration using the equation $v_f^2=v_i^2+2ad$. Since it departs from rest we have $a=\frac{v_f^2}{2d}$.

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have $F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1$

And we had the work done by force 2 as:

$W_2=F_2.d=\frac{mv_f^2}{2}-F_1d$

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

$W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J$

Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:

$W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}$

Which leads us to the previous equation straightforwardly.