Two forces, of magnitudes F_1 = 75.0 N and F_2 = 20.0 N, act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. Initially, the center of the block is at position x_i = -1.00 cm. At some later time, the block has moved to the right, and its center is at a new position, x_f = 2.00 cm.
Find the work W_1 done on the block by the force of magnitude F_1 = 75.0 N as the block moves from x_i = -1.00 cm to x_f = 2.00
Do the same for W_1 by force of magnitude F_2 = 20
W= F d
Since it's already parallel no need for cos
The Attempt at a Solution
Well, apparently I found out that F * d does not give the correct answer. I also tried subtracting one force or one resultant work from the other but it still isn't correct. The hint that is given to us has something to do with vectors and angles but I don't see how that will apply since the force is already parallel to the surface distance. Is there something hidden that I'm missing? Any thoughts?