A light spring having a force constant of 145 is used to pull a 7.50 sled on a h

Question

A light spring having a force constant of 145 is used to pull a 7.50 sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.30 ?

By how much does the spring stretch if it pulls on the sled horizontally?

By how much does the spring stretch if it pulls on the sled at 25.0 above the horizontal?

Explanation / Answer

By how much does the spring stretch if it pulls on the sled horizontally? The equation of motion is F - Fx = 0 Where F = k * d Fx = m a Therefore we have k * d - ma = 0 ==> d = ma / k = 7.50 * 2.30 / 145 = 0.1189 m = 11.89 cm By how much does the spring stretch if it pulls on the sled at 25.0 above the horizontal? In this case the equation of motion is F - Fx = 0 where Fx = m a cos@ k*d - m a cos25 = 0 ==> d = m a cos25 / k = 0.1189 * cos25 = 0.1077m = 10.77 m

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