Question

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If a 0.4 kg object hanging from a spring stretches it by 0.75 m, then by how much will the spring be stretched (in m) if a 0.8 kg object is suspended from it?

Answer 1

Solution :

Form Hooke's law;

The force applied to stretch a spring is given by ;

F = - k x

Where; k is the spring constant. It is a constant for a spring.  "x" is the stretch in the spring.

In our case, weight of the object works as the force.

F = m g

Then;

m g = - k x (i)

Now;

For two different case; equation (i) gives;

$\dfrac{m_2 \ g}{m_1\ g}= \dfrac{k\ x_2}{k\ x_1}\\ \ \\ or\\ x_2 =x_1\ \dfrac{m_2}{m_1}\\ \text{GIVEN:}\\ m_1= 0.4kg \ \ \ \ \ \ \ \ \ \ \ \ \ m_2= 0.8kg\\ x_1= 0.75 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_2=??\\ \text{Plugging in these values in the above equation, we get;}\\ x_2= 0.75\times\dfrac{0.8kg}{0.4kg}\\ x_2 = 1.5m$