A 53 g ice-cube can slide without friction up and down a 27 degree slope. The ic

Question

A 53 g ice-cube can slide without friction up and down a 27 degree slope. The ice-cube is pressed against a spring at the bottom of the slope, comressing the spring 10 cm. The spring constant is 25 N/m. When the ice-cube is released, what distance will it travel up the slope before reversing direction?

Ok, I think I have this but I would really appreciate confirmation of that!

W = 1/2*k*x^2, so .5 * .10^2 * 25 = 1.25.
Then, W = F * distance, so have to find force
Force = .053 *9.8 * sin 27, plug back in and get the distance = 5.3 m?

Explanation / Answer

mass of ice m = 53 g = 0.053 kg compression of the spring x = 10 cm =0.1 m spring constant k = 25 N / m from law of conservation of energy , maximum compressional PE of the spring = maximum gravitational energy        ( 1/ 2) kx^ 2= mgh from this height   h = kx^ 2 / ( 2mg )                             = 0.2406 m we know sin 27 = h / L from this required distance   L = h / sin 27                                               = 0.53 m          

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