Question

A block of 2 kg is suspended from a ceiling by a massless spring of spring constant k = 100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?

Answer 1

Answer:

If another 1 kg is added to the block the further elongation would be 0.1m

Explanation:

given :

mass=m=2kg

spring constant=k= 100 N/m

Let x be the spring elongation

From the free-body diagram,

kx = mg

x=mg/k

=2×9.8/100

=19.6/100

=0.196

≈0.2 m

Suppose, further elongation, when the 1 kg block is added ∆x.

Then,

k(x+∆x)=m’g

k  ∆x= 3g − 2g = g

⇒∆x=g/k

=9.8/100=0.098≈0.1 m

Answer 2

Answer:

A block of 2 kg is suspended from a ceiling by a massless spring of spring springnt है है है इस में में67 I I will follow up with you to know if you want to know if you want k = 100 N/m. What is the elongation of the spring? If another 1 kg is added to the block, what would be the further elongation?