7. What does the slope of a position-time graph represent?
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Answer:
Check attachment for free body diagram of the question.
I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.
Explanation:
Let frictional force be Fr acting down the plane
Let analyze the structure before inserting values
Using Newton's second law along the y-axis
ΣFy = Fnet = m•ay
Since the body is not moving in the y-direction, then ay = 0
N+PSinβ — WCosθ = 0
N+PSin20—441.45Cos15 = 0
N+PSin20—426.41 = 0
N = 426.41 — PSin20 , equation 1
The maximum Frictional force to be overcome is given as
Fr(max) = μsN
Fr(max) = 0.25(426.41 — PSin20)
Fr(max)= 106.6 —0.25•PSin20
Fr(max) = 106.6 — 0.08551P, equation 2
This is the maximum force that must be overcome before the body starts to move
Using Newton's law of motion in the x direction
Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward
Fnetx = ΣFx
Fnetx = P•Cosβ —W•Sinθ — Fr
Fnetx = P•Cos20—441.45•Sin15—Fr
Fnetx = 0.9397P — 114.256 — Fr
Equation 3
When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving
a. When P = 0
From equation 2
Fr(max) = 106.6 — 0.08551P
Fr(max) = 106.6 — 0.08551(0)
Fr(max)= 106.6 N
So, 106.6N is the maximum force to be overcome
So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.
Wx = WSinθ
Wx = 441.45× Sin15
Wx = 114.256 N.
Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.
So, finding the magnitude of frictional force
From equation 1
N = 426.41 — PSin20 , equation 1
N = 426.41 N, since P=0
Then, using law of kinetic friction
Fr = μk • N
Fr = 0.22 × 426.41
Fr = 93.81 N.
b. Now, when P = 190N
From equation 2
Fr(max) = 106.6 — 0.08551(190)
Fr(max) = 106.6 —16.2469
Fr(max)= 90.353 N
So, 90.353 N is the maximum force to be overcome
Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight
Fnetx = P•Cosβ —W•Sinθ
Fnetx = 190Cos20 — 441.45Sin15
Fnetx = 64.29N
So the force moving the body up the incline plane is 64.29N
Fnetx < Fr(max)
Then, the frictional force has not being overcome yet.
Then, the body is in equilibrium.
Then, applying equation 3.
Fnetx = 0.9397P — 114.256 — Fr
Fnetx = 0, since the body is not moving
0 = 0.9397(190) —114.246 — Fr
Fr = 64.297 N
Fr ≈ 64.3N
c. When, P = 268N
From equation 2
Fr(max) = 106.6 — 0.08551(268)
Fr(max) = 106.6 —16.2469
Fr(max)= 83.68 N
So, 83.68 N is the maximum force to be overcome
Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight
Fnetx = P•Cosβ —W•Sinθ
Fnetx = 268Cos20 — 441.45Sin15
Fnetx = 137.58 N
So the force moving the body up the incline plane is 137.58 N
Fnetx > Fr(max)
Then, the frictional force has being overcome.
Then, the body is not equilibrium.
So, finding the magnitude of frictional force
From equation 1
N = 426.41 — 268Sin20 , equation 1
N = 334.75 N, since P=268N
Then, using law of kinetic friction
Fr = μk • N
Fr = 0.22 × 334.75
Fr = 73.64 N
d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.
So, Fnetx = Fr(max)
Px — Wx = Fr(max)
From equation 1
Fr(max) = 106.6 — 0.08551P,
P•Cosβ-W•Sinθ = 106.6 — 0.08551P
P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P
P•Cos20—114.256=106.6 - 0.08551P
PCos20+0.08551P =106.6 + 114.256
1.025P=220.856
P = 220.856/1.025
P = 215.43 N
