You don't say which force acts in the direction of motion and which does not. Presumably F1, which is larger, is along the direction of motion. Otherwise the block would move backwards.
One force will do positive work and the other, negative work.
For (c), add them.
For (d) Kf - Ki equals the net work done (since there is no friction)
Do not expect step by step solutions here. We prefer to provide quidelines so you can do the work yourself.
Two forces, of magnitudes F1= 80.0N and F2= 50.0N, 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 x1= -3.00cm. At some later time, the block has moved to the right, and its center is at a new position, x2= 2.00cm.
a)Find the work W1 done on the block by the force of magnitude F1= 80.0N as the block moves from x1= -3.00 cm to x2= 2.00 cm
b)Find the work done by the force of magnitude F2 = 50.0 N as the block moves from x1= -3.00 Ncm to x2= 2.00cm.
c)What is the net work Wnet done on the block by the two forces?
d)Determine the change in the kinetic energy (Kf-Ki) of the block as it moves from x1= -3.00 cm to x2= 2.00 cm.
3 answers
a) F_1=80*(0.03+0.02)
b) F_2=50*(0.03+0.02)
c) answer part a + answer part b
d) same as part c
b) F_2=50*(0.03+0.02)
c) answer part a + answer part b
d) same as part c
a) 4