For a problem I have this: During a curling match, a 21.0 kg stone slides along

Question

For a problem I have this:
During a curling match, a 21.0 kg stone slides along a uniformlysmooth and level sheet of ice. The stone stops in the middle of thecenter circle, 40.8m from the release point. Take Mk=0.050 betweenthe stone and ice.

for the third part it asks:
The stone is swung in a circular arc of radius 0.70m just beforebeing released onto the ice. What force must be exerted by thethrowers arm at the bottom of the arc just before the stone isreleased?

I know I can use F=ma and then since theres a radius i can usef=mv2/r

but I don't know what forces there are besides mg..

any help?

Explanation / Answer

Wf = m g x    is the workdone against friction by the stone This must equal the kinetic energy of the stone when it wasreleased m g x = 1/2 m v2   willgive us the speed of the stone when it was released v = (2 m g x) = (2 * .05 * 9.8 *40.8) = 6.32 m/s Then the centripetal force that must be exerted on the stoneat release is F = m v2 / R = 21 * 6.322 / .7 = 1200N

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