The entire and thorough solution is included.
A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above
. If FP remains constant but the angle θ is increased, which of the following is true at some later time?
A.The force of friction between the block and surface will increase.
B.The force of friction between the block and surface will decrease.
C.The weight of the block will increase.
D.The weight of the block will decrease.
A.The force of friction between the block and surface will increase.
B.The force of friction between the block and surface will decrease.
C.The weight of the block will increase.
D.The weight of the block will decrease.
2 answers:
Response:
B. The friction force between the block and surface will diminish.
Clarification:
The force of friction can be expressed as
where represents the friction coefficient and
signifies the normal force.
As the student pulls the block with force at an angle
, the normal force acting on the block subsequently becomes
and therefore, the frictional force changes to
.
With an increase in , the
increases, which subsequently reduces the normal force
, indicating a decrease in the frictional force; hence choice B is correct.
Note: While choice D seems appealing, it is incorrect since the weight remains unaffected by the external forces acting on the block.
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Answer:
a = 18.28 ft/s²
Explanation:
the values provided are:
duration of force application, t= 10 s
Work done = 10 Btu
mass of the object = 15 lb
acceleration, a =? ft/s²
1 Btu = 778.15 ft.lbf
thus, 10 Btu = 7781.5 ft.lbf
m = 0.466 slug
So,
the work is equivalent to the change in kinetic energy
The acceleration of the object is therefore
a = 18.28 ft/s²
the constant acceleration of the object is calculated to be 18.28 ft/s²
assuming north-south is along the Y-axis and east-west along the X-axis
X = total X-displacement
from the graph, total displacement in the X-direction is computed as
X = 0 - 20 + 60 Cos45 + 0
X = 42.42 - 20
X = 22.42 m
Y = total Y-displacement
from the graph, total displacement in the Y-direction is computed as
Y = 40 + 0 + 60 Sin45 + 50
Y = 90 + 42.42
Y = 132.42 m
To calculate the magnitude of the net displacement vector, we apply the Pythagorean theorem, yielding
magnitude: Sqrt(X² + Y²) = Sqrt(22.42² + 132.42²) = 134.31 m
Direction: tan⁻¹(Y/X) = tan⁻¹(132.42/22.42) = 80.4 deg north of east
Response:
Clarification:
We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:
Where v signifies the number of gallons emptied from the tub.
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