Max Height for Pole-Vaulter with Speed 6.00 m/s

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The discussion focuses on calculating the maximum height a pole-vaulter can achieve with a speed of 6.00 m/s, emphasizing the use of energy conservation principles. The kinetic energy from the vaulter's speed is converted into gravitational potential energy to determine the height. Additionally, there is a separate inquiry about calculating the average power expended by a person making homemade ice cream, involving the force applied and the circular motion of the crank handle. The key to solving both problems lies in understanding the relationships between kinetic energy, potential energy, work, and power. Energy conservation and work calculations are central to finding the required solutions.
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A pole-vaulter approaches the takeoff point at a speed of 6.00 m/s. Assuming that only this speed determines the height to which he can rise, find the maximum height at which the vaulter can clear the bar.
in this 1 I am lost......
I ONLUY GOT 1 GIVINGS (6.00 m/s )
should i use the ...h0-hf = m ??


person is making homemade ice cream. She exerts a force of magnitude 19 N on the free end of the crank handle, and this end moves in a circular path of radius 0.32 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.9 s, what is the average power being expended?


i know that the P= WORK/TIME...
AND P= CHANGE IN ENERGY/ TIME

I TRIDE BOTH ..BUT NONE OF THEM WORKED FOR ME...
 
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iceT said:
A pole-vaulter approaches the takeoff point at a speed of 6.00 m/s. Assuming that only this speed determines the height to which he can rise, find the maximum height at which the vaulter can clear the bar.
in this 1 I am lost......
I ONLUY GOT 1 GIVINGS (6.00 m/s )
should i use the ...h0-hf = m ??
Think in terms of energy conservation. The vaulter starts with some kinetic energy due to his speed; assume that that energy gets converted to gravitational potential energy to find the maximum height.


person is making homemade ice cream. She exerts a force of magnitude 19 N on the free end of the crank handle, and this end moves in a circular path of radius 0.32 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.9 s, what is the average power being expended?
For every complete turn of the crank, how much work is done? Once you figure out the work, then compute the power.
 
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