Question

In 0.601 s, a 13.1-kg block is pulled through a distance of 4.19 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 494 N/m. By how much does the spring stretch

Answer 1

Answer:

0.615 m

Explanation:

We need to determine the force on the spring first. By Newton's second law of motion, force is the product of the mass and acceleration. The mass is given.

The acceleration is determined using the equation of motion.

Given parameters:

Initial velocity, u = 0.00 m/s

Distance, s = 4.19 m

Time, t = 0.601 s

We use the equation

$s = ut+\frac{1}{2}at^2$

With u = 0.00 m/s,

$s = \frac{1}{2}at^2$

$a = \dfrac{2s}{t^2}$

$a = \dfrac{2\times4.19\text{ m}}{(0.601\text{ s})^2} = 23.2\text{ m/s}^2$

The force is

$F = (13.1\text{ kg})(23.2\text{ m/s}^2) = 303.92 \text{ N}$

From Hooke's law, the extension, e, of a string is given by

$e = \dfrac{F}{k}$

where k is the spring constant.

Hence,

$e = \dfrac{303.92\text{ N}}{494\text{ N/m}} = 0.615\text{ m}$