Question

An unstretched spring hangs from the ceiling with a length of 0.95 m. A 7.9 kg block is hung from the spring and the spring stretches to be now 1.06 m long. How long will the spring be if a 1.3 kg block is hung from the spring? For both cases the block is placed gently are there are vibrations of the spring are allowed to settle before any measurements are made.

Answer 1

Answer:

The distance is 0.9681 m.

Explanation:

Given that,

Length = 0.95 m

Mass of block = 7.9 kg

Stretches spring length = 1.06 m

another mass = 1.3 kg

Magnitude of restoring force acting in the spring equal to the magnitude of applied force

$F_{rest}=F_{ap}$

$F = mg$

We need to calculate the spring constant

Using Hook's law,

$F_{r}=kx$

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

Put the value into the formula

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{7.9\times9.8}{1.06-0.95}$

$k=703.81\ N/m$

We need to calculate the distance

Using formula of spring constant

$x=\dfrac{mg}{k}$

Put the value into the formula

$x=\dfrac{1.3\times9.8}{703.81}$

$x=0.0181\ m$

We need to calculate the total distance

Using formula of distance

$d=x+l$

$d=0.0181+0.95$

$d=0.9681\ m$

Hence, The distance is 0.9681 m.