Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizont
ally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?
1 answer:
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Answer:
ω = 2.1 rad/sec
Explanation:
- As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
- This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
- Now, we need to ask ourselves: what supplies this force?
- In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
- We know that this force can be expressed as follows:
where μs = coefficient of static friction between the rock and the merry-
go-round surface = 0.7, and Fn = normal force.
- In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
- Fn = m*g (2)
- This static friction force is just the same as the centripetal force.
- The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:
- Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:
- This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As
As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So