Question

In 0.820 s, a 7.20-kg block is pulled through a distance of 3.85 m on a frictionless horizontal surface, starting from rest. The block has a constant acceleration and is pulled by means of a horizontal spring that is attached to the block. The spring constant of the spring is 390 N/m. By how much does the spring stretch?

Answer 1

We will first find the acceleration of the block using the information given. Given that the block starts from rest, its initial velocity is zero, and hence distance s covered in time t is where a is the acceleration. Hence .

Force on the block by Newton's second law, where m is the mass of the block. Using m=7.2kg as given, we find F=7.2*11.45=82.45N.

This force is transferred to the block by the spring. If the spring extends by x relative to its rest length, it exerts a force on the block (and on the agent pulling the block). Hence, equating the forces, we find .

    kx -----> spring <-----kx

block--------------------------- agent----->F

Hope the answer was helpful!