On earth a block has a weight of 88 N. This block is sliding on a horizontal surface on the moon, where the acceleration due
to gravity is 1.60 m/s
2. As Figure 4.38
a shows, the block is being pulled by a horizontal rope in which the tension is
T = 24 N. The coefficient of kinetic friction between the block and the surface is μ
k = 0.20. Determine the acceleration of the block.
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| Figure 4.38 |
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(a)
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A block is sliding on a horizontal surface on the moon. The tension in the rope is  .
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(b)
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The free-body diagram for the block, including a kinetic frictional force  .
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Concept Questions and Answers |
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Which of Newton's laws of motion provides a way to determine the acceleration of the block?
Answer:
Newton's second law allows us to calculate the acceleration as ax = ΣFx/m, where ΣFx is the net force acting in the horizontal direction and m is the mass of the block.
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This problem deals with a situation on the moon, but the block's mass on the moon is not given. Instead, the block's earth-weight
is given. Why can the earth-weight be used to obtain a value for the block's mass that applies on the moon?
Answer:
Since the block's earth-weight Wearth is related to the block's mass according to Wearth = mgearth, we can use Wearth = 88 N and gearth = 9.80 m/s2 to obtain m. But mass is an intrinsic property of the block and does not depend on whether the block is on the earth or on the moon.
Therefore, the value obtained for m applies on the moon as well as on the earth.
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Does the net force ΣFx equal the tension T?
Answer:
No. The net force ΣFx is the vector sum of all the external forces acting in the horizontal direction. It includes the kinetic frictional force
fk as well as the tension T.
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Solution
Figure 4.38
b shows the free-body diagram for the block. The net force along the
x axis is Σ
Fx = +
T -
fk, where
T is the magnitude of the tension in the rope and
fk is the magnitude of the kinetic frictional force. According to Equation 4.8,
fk is related to the magnitude
FN of the normal force by
fk = μ
kFN, where μ
k is the coefficient of kinetic friction. The acceleration
ax of the block is given by Newton's second law as
We can obtain an expression for
FN by noting that the block does not move in the
y direction, so
ay = 0 m/s
2. Therefore, the net force Σ
Fy along the
y direction must also be zero. An examination of the free-body diagram reveals that Σ
Fy = +
FN -
mgmoon = 0, so that
FN =
mgmoon. The acceleration in the
x direction becomes
Using the earth-weight of the block to determine its mass, we find
The acceleration of the block is, then,
