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 maximum kinetic energy at any time in its motion? Neglect friction. A. 1 J B. 0.025 J C. 0.05 J D. Not enough information given E. 0.1 J

Answer 1

Answer:

C. 0.05 J

Explanation:

Given that

Spring constant ,K= 10 N/m

Distance ,x = 0.1 m

The amplitude ,A= 0.1 m

The stored energy potential energy in the spring

$U=\dfrac{1}{2}KA^2$

Now by putting the values

$U=\dfrac{1}{2}\times 10\times 0.1^2\ J$

U=0.05 J

Now From energy conservation

The maximum kinetic energy = The Maximum potential energy

KE= U

KE= 0.05 J

Therefore the answer will be C.

C. 0.05 J