Expert: Daniel Mazur - 10/2/2008

Question
If a 2kg block is moving on a frictionless surface at 7m/s and collides with a spring with force constant 1120N/m, how far will the spring compress.  How can I calculate the force the block imparts on the spring?  Since there is no friction the block has no acceleration, so doing F=ma is useless.  Can you please help.

Answer
Hello Will,
There are more answers possible (different in the level of detail and methods used) depending on your level. I will assume you are at high school, please correct me in a follow-up if I am mistaken.
As there is no friction, the block looses no energy until it touches the spring. The spring had been until that moment in a relaxed state. As you have not been told the length of the spring, we assume it long enough to completely stop the block.
The key equation here is the spring force equation, which is F=kx, where k=1120N/m. As x is the length difference between the compressed (or elongated, generally) and the relaxed spring. In other words, if you compress the spring by 'x' from its relaxed state, you will be using a force F=kx.
I think that high school physics does not solve this problem as "Calculate the force the block exerts on the spring *at any time t*". The most high school physics does is calculate the force at the moment, when the block's velocity is zero. The latter can be calculated using the conservation of energy law: The kinetic energy (1/2)mv0^2 of the block before impact must be then exactly equal to the potential energy stored in the spring (1/2)kx^2, when the block is stationary. The end position is then x_stop=sqrt(m/k)*v0 and the force at that point is F(x_stop)=sqrt(k*m)*v0. You can easily draw a graph of force vs. position x now: it is linear, originating in [0,0] point, it has slope 'k' (F=kx) and the end point is [x_stop,F(x_stop)].
As there is only one interaction in this problem (spring and block), the force of the spring on the block and the force of the block on the spring are of equal magnitude and opposite orientation (Newton's 3rd law of motion).
So, here you are. If you need to know more, please send me a followup question.
Cheers,
Daniel