Two forces, of magnitudes F 1 = 100 N and F 2 = 35.0 N , act in opposite directi

Question

Two forces, of magnitudes F1 = 100 N and F2 = 35.0 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 = -5.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 2.00 cm .

a) Find the work W1 done on the block by the force of magnitude F1 = 100 N as the block moves from xi = -5.00 cm to xf = 2.00 cm .

b) Find the work W2 done by the force of magnitude F2 = 35.0 N as the block moves from xi = -5.00 cm to xf = 2.00 cm .

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

d)Determine the changeKfKi in the kinetic energy of the block as it moves from xi = -5.00 cm to xf = 2.00 cm .

Explanation / Answer

a) Work (energy) can be found from force x distance.

So for the first one: F x d
100 N * 7cm * 1m/100cm = 7.0 J

b) The next is the same, but since the force is acting in the opposite direction, it gets an opposite sign.
-35 N * 7m * 1m/100cm = - 2.45JJ

c) Wnet = 7J - 2.45 J = 4.55 J

For (d), you could either assume that it's identical to (c), since there is no friction. Or you could compute one from the forces and see that it's the same. Assume it starts at v=0, and has mass of z kg.
F(net) = 100 - 35 = 65N
a = F/(z kg) = 65N/(z kg)

v^2 = 2 a d
v^2 = 2 * 65N/(z kg) * 7cm
v^2 = 910 N m / (zkg)

KE = .5 * m * v^2
KE = .5 * zkg * 910 N m / zkg
KE = .5 * 910 N m
KE = 4.55 J

Since the surface is frictionless, no work is lost to friction. Since the block has not changed elevation, its gravitational potential energy is unchanged. Therefore, all of the work done went into kinetic energy hence the answer to D

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