Question

A light spring has a force constant 20 N/m. A body
of mass 2 kg hanging in equilibrium from the spring
is pulled down by 10 cm and released. Subsequently
the maximum kinetic energy of the body will be
(1) 0.2 J
(2) 0.4
(3) 0.1 J
(4) 2J​

Answer 1

Solution :

⏭ Given:

✏ Force constant of a spring = 20 N/m

✏ Mass of a body = 2 kg

✏ Change in length of spring = 10 cm

⏭ To Find:

✏ Maximum kinetic energy of the body.

⏭ Formula:

✏ Formula of maximum kinetic energy stored in spring in terms of spring constant and change in length of spring is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{KE = \frac{1}{2} k {x}^{2} }}}}} \: \star$

⏭ Terms indication:

✏ KE denotes kinetic energy

✏ k denotes spring constant

✏ x denotes change in length of spring

⏭ Conversation:

✏ 10 cm = 0.1 m

⏭ Calculation:

$\mapsto \sf \: KE = \frac{1}{2} \times 20 \times {0.1}^{2} \\ \\ \mapsto \sf \: KE = 0.5 \times 20 \times 0.01 \\ \\ \star \: \boxed{ \bold{ \sf{ \red{KE = 0.1 \: J}}}} \: \star$

⏭ Additional information:

• Spring force is conservative type of force.

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Given ,

• Spring or force constant (k) = 20 N
• Mass of the body (m) = 2 kg
• Compression (x) = 10 cm or 0.1 m

We know that , kinetic energy of spring is given by

$\large \mathtt{ \fbox{Kinetic \: energy = \frac{1}{2}k {(x)}^{2} }}$

Thus ,

$\sf \hookrightarrow Kinetic \: energy = \frac{1}{2} \times 20 \times {(0.1)}^{2} \\ \\ \sf \hookrightarrow Kinetic \: energy = 10 \times \frac{1}{100} \\ \\ \sf \hookrightarrow Kinetic \: energy = 0.1 \: \: joule$

Hence , the kinetic energy of the spring is 0.1 joule

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