Question

A block of mass m=25 kg moves on a smooth horizontal surface with a velocity $\overrightarrow{v}=3m{s}^{-1}$ meets the spring of spring constant $K=100N/m$ fixed at one end as shown in figure. The maximum compression of the spring and velocity of block as it returns to the original position respectively are

• A. $1.5m,-3m{s}^{-1}$
• B. $1.5m,0m{s}^{-1}$
• C. $1.0m,3m{s}^{-1}$
• D. $0.5m,2m{s}^{-1}$

Answer 1

Answer: (A)

$1.5m,-3m{s}^{-1}$

According to law of conservation of energy, when the block strikes the spring the kinetic energy of block is converted to potential energy of spring.

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

where $x$ is compression of spring.

$x=\sqrt{\frac{m{v}^2}{k}}$

$x=\sqrt{\frac{25\times 3^2}{100}}$

$x=1.5m$

When the block returns to the original position, again the potential energy is converted into the kinetic energy. Hence, velocity of block is same as before but sign changes because it returns to mean position

So, $v=-3m/s$