θ=sin−1(25N+40N(10kg)(10ms2))=sin−1(0.65)
θ=40.5∘
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Tension : Example Question #4
Consider the following system:
Slope_2
If the coefficient of static friction is 0.25, the angle measures 30∘, the force of tension is 20N, and the block is motionless, what is the mass of the block?
g=10ms2
Possible Answers:
3.2kg
12.4kg
8.9kg
7.1kg
5.4kg
Correct answer:
7.1kg
Explanation:
There are three relevant forces acting on the block in this scenario: tension, friction, and gravity. We are given tension, so we will need to develop expressions for friction and gravity. Since the block is motionless, we can say:
Fnet=0
Fg−Ff−T=0
Plugging in expressions for the force of gravity and friction, we get:
mgsin(θ)−μsmgcos(θ)−T=0
Rearranging for the mass, we get:
m=Tg(sin(θ)−μscos(θ))
We know all of these values, allowing us to solve:
m=20N(10ms2)(sin(30∘)−(0.25)cos(30∘))
m=7.1kg
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Tension : Example Question #5
Consider the following system:
Slope_2
If the block has a mass of 5kg and the angle measures 20∘, what is the minimum value of the coefficient of static friction that will result in a tension of 0N?
g=10ms2
Possible Answers:
0.40
0.27
0.45
0.11
0.36
Correct answer:
0.36
Explanation:
Since there is no tension, there are only two relevant forces acting on the block: friction and gravity. Since the block is motionless, we can also write:
Fnet=0
Fg−Ff=0
Fg=Ff
Substitute the expressions for these two forces:
mgsin(θ)=μsgcos(θ)
Canceling out mass and gravitational acceleration, and rearranging for the coefficient of static friction, we get:
μs=sin(θ)cos(θ)=tan(θ)
μs=0.36
θ=sin−1(25N+40N(10kg)(10ms2))=sin−1(0.65)
θ=40.5∘
Report an Error
Tension : Example Question #4
Consider the following system:
Slope_2
If the coefficient of static friction is 0.25, the angle measures 30∘, the force of tension is 20N, and the block is motionless, what is the mass of the block?
g=10ms2
Possible Answers:
3.2kg
12.4kg
8.9kg
7.1kg
5.4kg
Correct answer:
7.1kg
Explanation:
There are three relevant forces acting on the block in this scenario: tension, friction, and gravity. We are given tension, so we will need to develop expressions for friction and gravity. Since the block is motionless, we can say:
Fnet=0
Fg−Ff−T=0
Plugging in expressions for the force of gravity and friction, we get:
mgsin(θ)−μsmgcos(θ)−T=0
Rearranging for the mass, we get:
m=Tg(sin(θ)−μscos(θ))
We know all of these values, allowing us to solve:
m=20N(10ms2)(sin(30∘)−(0.25)cos(30∘))
m=7.1kg
Report an Error
Tension : Example Question #5
Consider the following system:
Slope_2
If the block has a mass of 5kg and the angle measures 20∘, what is the minimum value of the coefficient of static friction that will result in a tension of 0N?
g=10ms2
Possible Answers:
0.40
0.27
0.45
0.11
0.36
Correct answer:
0.36
Explanation:
Since there is no tension, there are only two relevant forces acting on the block: friction and gravity. Since the block is motionless, we can also write:
Fnet=0
Fg−Ff=0
Fg=Ff
Substitute the expressions for these two forces:
mgsin(θ)=μsgcos(θ)
Canceling out mass and gravitational acceleration, and rearranging for the coefficient of static friction, we get:
μs=sin(θ)cos(θ)=tan(θ)
μs=0.36
