In 0.750 s, a 7.00 kg block is pulled through a distance of 4.00 m on a friction

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? 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?

Explanation / Answer

Given,

t = 0.75 m

d = 4 m

m = 7 kg

k = 415 N/m

So,

let a is the acceleration of the block

d = vo*t + 0.5*a*t^2

d = 0 + 0.5*a*t^2

a = 2*d/t^2

= 2*4/0.75^2

= 14.22 m/s^2

So,

Force exerted by the spring,

= F = m*a

= 14.22*7

= 99.54 N

Then,

From Hook's law,

extension of the spring,

= x= F/K

= 99.54/415

= 0.239 m

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