Two forces, of magnitudes F 1 = 100 N and F 2 = 50.0 N , act in opposite directi
ID: 1365085 • Letter: T
Question
Two forces, of magnitudes F1 = 100 N and F2 = 50.0 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 = -4.00 cm . At some later time, the block has moved to the right, and its center is at a new position, xf = 5.00 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 = 5.00 cm .
Express your answer numerically, in joules.
Part B
Find the work W2 done by the force of magnitude F2 = 50.0 N as the block moves from xi = -4.00 cm to xf = 5.00 cm .
Express your answer numerically, in joules.
Part C
What is the net work Wnet done on the block by the two forces?
Express your answer numerically, in joules.
Part D
Determine the changeKfKi in the kinetic energy of the block as it moves from xi = -4.00 cm to xf = 5.00 cm .
Express your answer numerically, in joules.
Explanation / Answer
here,
as work done = force acting * displacement *cos(theta)
theta is the angle between force and displacement vector
A)
the work W1 done on the block by the force of magnitude F1 = 100 N , W1 = F1*( 5 - ( -4))*10^-2 * cos(90)
W1 = 9 J
the work W1 done on the block by the force of magnitude F1 = 100 N is 9 J
B)
the work W2 done by the force of magnitude F2 = 50.0 N as the block moves , W2 = F2 * ( 5 - ( - 4)) * 10^-2 * cos(180)
W2 = - 4.5 J
the work W2 done by the force of magnitude F2 = 50.0 N as the block moves is -4.5 J
C)
the net work Wnet done on the block by the two forces , W = W1+ W2
W = 4.5 J
D)
using work energy theorm
work done = change in kinetic energry
the changeKfKi in the kinetic energy of the block as it moves from xi = -4.00 cm to xf = 5.00 cm is 4.5 J
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