Question

Two forces, of magnitudes F1 = 100 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface. Initially, the center of the block is at position xi = -4.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 1.00 cm .

Part A

Find the work W1 done on the block by the force of magnitude F1 = 100 N as the block moves from xi = -4.00 cm to xf = 1.00 cm .

Part B

Find the work W2 done by the force of magnitude F2 = 45.0 N as the block moves from xi = -4.00 cm to xf = 1.00 cm .

Part C

What is the net work Wnet done on the block by the two forces?

Part D

Determine the changeKf−Ki in the kinetic energy of the block as it moves from xi = -4.00 cm to xf = 1.00 cm .

Answer 1

F₁=100 N and F₂=-45 N where - sign indicates forces are in  opposite directions

work W1 done on the block by the force of magnitude F1 = 100 N is W1 =F*d*sin( $\theta$ )

where $\theta$ =angle between force and displacement and d=displacement

=>W1= 100*(0.01-(0.04)) =5J ...................part a

b) W2 =-45*(0.01-(0.04))=-2.25J

c)net work done Wnet =W1+W2=5-2.25 =2.75J

or Wnet =Fnet*d =(100-45)*0.05 =2.75J

d) work done on block converted to kinetic energy of block

=>K.E_i+W =K.E_f =>changeKf−Ki in the kinetic energy of the block =W=2.75J