A 2.0 kg block is released from rest on a frictionless 30 degree incline.

[joanneleeee]

Homework Statement

A 2.0 kg block is released from rest on a frictionless 30 degree incline. After sliding 0.80 m it comes in contact with the spring, and after sliding an additional 0.20 m, it comes momentarily to rest. What is the spring constant?

I knew that spring constant is equal to the force it takes to compress a spring over the distance the spring compresses. I thought that the force would just be mg which is 2.0x9.8=19.6, and I just divided that value by 0.20 to get 98 N/m, but my teacher told me that that isn't the correct answer...

 

[Poley]

I think that the slope of the incline must be considered. The force that the block applies to the spring is not [itex]mg[/itex] but [itex]mg \cos(\theta)[/itex], so the spring constant is [itex]k=\frac{mg \cos(\theta)}{x}[/itex].

Hopefully this will provide you with the correct answer!

 

[Steely Dan]

I knew that spring constant is equal to the force it takes to compress a spring over the distance the spring compresses. I thought that the force would just be mg which is 2.0x9.8=19.6, and I just divided that value by 0.20 to get 98 N/m, but my teacher told me that that isn't the correct answer...

Strictly speaking, the spring constant is equal to the force exerted by the spring divided by the compression distance. What you said is only true if the external force causes the object to reach a stationary position, for then the spring force would exactly balance the external force and thus the two would be equal in magnitude. But this is not an equilibrium situation; right afterward, the spring would uncompress and start pushing the object back up the incline.

In these situations you should use a conservation of energy approach. The initial gravitational potential energy of the object is converted fully into elastic potential energy of the spring at the moment of maximum compression.