When Push Comes to Shove?

Question

Two forces, of magnitudes F_1 = 80.0 N and F_2 = 30.0 N, 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 x_i = -2.00 cm. At some later time, the block has moved to the right, and its center is at a new position, x_f = 1.00 cm.

A) Find the work W_1 done on the block by the force of magnitude F_1 = 80.0 N as the block moves from x_i = -2.00 cm to x_f = 1.00 cm.

B) Find the work W_2 done by the force of magnitude F_2 = 30.0 N as the block moves from x_i = -2.00 cm to x_f = 1.00 cm.

C) What is the net work  done on the block by the two forces?

D) Determine the change K f - K i in the kinetic energy of the block as it moves from x_i = -1.00 cm to x_f = 1.00 cm.

Anonymous · Accepted Answer

The particle moved in the positive direction, so I will assume F1 (the bigger force) is positive and F2 is negative.  If the particle started out moving really fast to the right, this could be wrong, but that's what I'll assume.

a) work = force * distance

b) work = force * distance--in this case, though, F2 is negative, so the work is negative.

c)  add em up.  OR  you could calculate the net force and use work = netforce * distance

d)  The kinetic energy increases as work is done.  delta KE = work = netforce*distance.  TAKE CARE--they changed the distance, so the answer isn't the same as part c.

Grant d · Answer

a) Work = F*D = 80*(1--2) = 80* 3 = 240 N.cm
b) Work = F*D = 30(-2-1) = 30 * -3 = -90 N.cm
c) Net Work = 240-90 = 150 N.cm
d) KE = Work = 150 N.cm

Randy C · Answer

A & B)     W=F(xi-xf)

C)              Wnet = Fnet(xi - xf)

D)              Wnet = mvf2 - mvI2 = Kf - Ko
                   Implying that        Wnet = ΔK 

I think you can handle the math.

Anonymous · Answer

A)W=80X.03=2.40J
B)W=30X.03=0.90J
C)W=3.30J
D)Change in K.E.=workdone=3.30J

When Push Comes to Shove?

4 Answers