Question

A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on the spring which stretches the spring and does 20 J of work. How far is the spring stretched as a result of the work done by the second mass?

Answer 1

Answer:

 x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm  = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

  k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

$W = \dfrac{1}{2}kx'^2$

$20= \dfrac{1}{2}\times 39.2\times x'^2$

 x'² = 1.0204

 x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.