A spring with a force constant of 5100 N/m and a rest length of 2.6 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m) ? (Assume the rock is launched from ground height.) m How fast is it going (in m/s) when it hits the ground? m/s

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Wally

Physics

20 November, 05:44

A spring with a force constant of 5100 N/m and a rest length of 2.6 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m) ? (Assume the rock is launched from ground height.) m How fast is it going (in m/s) when it hits the ground? m/s

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Kymani Carrillo · Answer 1

First let’s find the total force exerted on the rock.

Fs=1/2kx^2 = 1/2*5100 * (1.6) ^2=6528N

Then setting it equal to U=mgh and solving for h we get h=6528N / (9.8m/s^2*48kg) = 13.9m

Then using K=1/2mv^2 we can solve for the final velocity of the rock.

(6528N*2) / 48kg then taking the sqrt of that we get Vf = 16.5m/s