Find the work W1 done on the block by the force of magnitude F1 = 75.0 N as the block moves from xi = -1.00 cm to xf = 2.00 cm .?
Two forces, of magnitudes F1 = 75.0 N and F2 = 45.0 N , act in opposite directions on a block, which sits atop a frictionless surface, as shown in the figure. (Figure 1)Initially, the center of the block is at position xi = -1.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 2.00 cm .
- AshLv 71 month ago
F(net) = 75.0 - 45.0 = 30.0 N
displacement = xf - xi = 2.00 -(-1.00) = 3.00 cm = 0.0300 m
Work done on block = F(net) x displacement = (30.0 N)x (0.0300 m)= 0.900 J