Two forces, of magnitude F1=50.0 N and F2=22.0 N, act in opposite directions ona

Question

Two forces, of magnitude F1=50.0 N and F2=22.0 N, act in opposite directions ona block which sits atop a horizontal frictionless surface as shown. 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 position xf=+6.00 cm.

a) What is the work done by force F2 as the block moves from x=xi to x=xf?

b) What is the change in the block's kinetic energy as it moves from x=xi to x=xf?

Please show your work. I did this problem but am not sure where I went wrong. Thanks.

Explanation / Answer

F1 = 50 N F2 =22N

net force = F1-F2 = (50-22) = 28N

a ) work done by force F2 = -Fdx = -22 *(6+4) = -220J

b) change in kinetic energy of the block = work done

                   = Fnet *dx

                  = 28*(6+4) = 280J

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