Question

In 0.750 s, a 7.00-kg block is pulled through a distance of 4.00 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 415 n/m. By how much does the spring stretch at its maximum displacement?

Answer 1

2 answers · Physics

Best Answer

The acceleration of the block can be determined from:

d=v0*t + a*t^2/2 ; v0=0 => d=a*t^2/2

a=2d/t^2 = 14.22m/s^2

The force applied F=m*a

F=7*14.22 = 99.56N

Since the force of a spring F=k*x, x=elongation, k=spring constant

x=F/k

x=99.56/415=0.24m