How Far Will the Spring Compress When Dropping a 5.8 kg Block?

[DStan27]

A vertical spring with k = 495 N/m is standing on the ground. You are holding a 5.8 kg block just above the spring, not quite touching it.

How far does the spring compress if you let go of the block suddenly?

i thought i could just use k = F/distance

so 495(D) = (5.8)(9.8)

D = .1148 but this is not correct. i take it i am doing something wrong with my acceleration but I'm not sure what. can anyone help me with this?

 

[DStan27]

nevermind i figured it out.

 

[quicknote]

Your approach is correct, but you need to account for the weight of the block in your calculation. The force exerted by the spring on the block is equal to the weight of the block, which is given by mg, where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2). So the equation should be kx = mg, where x is the distance the spring compresses and mg is the weight of the block. Rearranging, we get x = mg/k. Plugging in the values, we get x = (5.8 kg)(9.8 m/s^2)/(495 N/m) = 0.1167 m. So the spring would compress approximately 0.1167 meters if you let go of the block suddenly.