A 0.5 kg block moving at a speed of 12 ms⁻¹ compresses a spring through a distance of 30 cm when its speed is halved, the spring constant of the spring will be ________ Nm⁻¹.

CONCEPT:

• According to the law of conservation of energy, the loss of kinetic energy is equal to the gain in potential energy and it is written as;

           Loss of kinetic energy = gain in potential energy.

• Kinetic energy is defined as energy possessed by the body due to the motion and it is written as;

           K.E. = \(\frac{1}{2}mv^2\)

          here we have m as the mass of the body, and v is the velocity.

• The potential energy of the body is written as;

           P.E. = \(\frac{1}{2}kx^2\)

          here we have k as the spring constant and x as the distance.

CALCULATION:

Given: mass, m = 0.5 kg, x = 30 cm = 0.3 m,  v = 12 m/s

According to the conservation of energy we have;

Loss of kinetic energy = gain in potential energy

Here let us suppose the initial velocity of the block v m/s  and when it is compressed its velocity is halved which is v/2, therefore,

loss of kinetic energy, K.E. = \(\frac{1}{2} mv^2 \) -  \(\frac{1}{2}m(\frac{v}{2})^2\)

Now, 

 \(\frac{1}{2} mv^2 \) - \(\frac{1}{2}m(\frac{v}{2})^2\) = \(\frac{1}{2}kx^2\)

⇒  \(\frac{1}{2} mv^2 \) - \(\frac{1}{2}m(\frac{v^2}{4})\) = \(\frac{1}{2}kx^2\)

⇒ \(\frac{3mv^2}{8} = \frac{1}{2}kx^2\)

⇒ k = \(\frac{3mv^2}{4x^2}\)

Now, on putting the given values we have;

k = \(\frac{3\times0.5 \times 12^2}{4 \times 0.3^2}\)

⇒k = 600 N/m

Hence spring constant k = 600 N/m.