Question

If a spring with a spring constant k = 10 N/m is stretched from its equilibrium position by 0.1 m, and released, what would be the maxi- mum kinetic energy at any time in its motion? Neglect friction.1. 0.025 J2. 0.1 J3. Not enough information given4. 0.05 J5. 1 J

Answer 1

Answer:4

Explanation:

Given

spring constant $k=10 N/m$

if it is stretched 0.1 m from equilibrium position

If the spring is at maximum extension then the Elastic Potential Potential energy is maximum and is equal to Total energy of the system

When the spring is at equilibrium , total energy will be equal to kinetic energy

So maximum kinetic Energy is equal to

$=\frac{1}{2}kx^2$

$=\frac{1}{2}\times 10\times (0.1)^2$

$=0.05\ J$

                     

Answer 2

Answer:

Option 4. 0.05 J

Solution:

As per the question:

Spring constant, k = 10 N/m

Equilibrium position, x = 0.1 m

Now,

The potential energy of the spring is given by:

$U = \frac{1}{2}kx^{2}$

And also from the principle of conservation of energy:

KE = U                             (1)

where

KE = Maximum Kinetic Energy

U = Potential energy

Thus

KE =  U = $\frac{1}{2}kx^{2}$

KE =  U = $0.5\times 10\times 0.1^{2} = 0.05\ J$