When a bow is stretched it stores 10j and the arrows mass is 50g as it is projected show that the arrow has a speed of 20m/s
1 answer:
Answer:
20 m/s
Explanation:
The stored energy is transformed to kinetic energy and since kinetic energy is given by
where m is the mass of arrow in Kg and v is the velocity of the arrow in m/s. From the principle of energy, all the energy is transformed to KE.
KE= Energy stored in bow
Changing mass from 50 g to Kg we divide by 1000 hence m=50/1000=0.05 Kg
Substituting 0.05 Kg for m and 10 J for KE then we have
0.5\times 0.05\times v^{2}=10 J
Making v the subject of the formula then
Therefore, the velocity is equivalent to 20 m/s
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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.