Answered

An object of mass 12.0 kg is placed on the pan of a spring-powered scale. The spring constant of the scale is 31.3 N/cm. What distance will the spring be compressed?

Verified Answer :

AbhiGhost

F= mg = 12x9.8 =117.6N. According to Hooke's law, F =kx. x =F/k =117.6/31.3 = 3.75cm. So compression in the spring is 3.75cm.

Answer:

Distance, x = 3.75 cm

Explanation:

Given that,

Mass of the object, m = 12 kg

Spring constant of the scale, k = 31.3 N/cm

To find,

The distance compressed by the spring.

Solution,

We know that Hooke's law is used to find the force and the distance compressed. Its mathematical expression is given by :

\(F=-kx\)

\(x=\dfrac{F}{k}\)

\(x=\dfrac{mg}{k}\)

\(x=\dfrac{12\ kg\times 9.8\ m/s^2}{31.3\ N/cm}\)

x = 3.75 cm

So, the spring will be compressed to a distance of 3.75 cm.

Related Questions