Question

A block with a mass of $1\text{kg}$ is connected to a massless spring with a force constant of $100\text{N/m}$ and is placed on a smooth horizontal surface. At $t=0$, a constant force of $F=10\text{N}$ is applied on the block, when the spring is in its natural length. The speed of the block, when it is displaced by $8\text{cm}$ from its mean position will be

• A. $80\text{cm/s}$
• B. $16\text{cm/s}$
• C. $60\text{cm/s}$
• D. $25\text{cm/s}$

Answer 1

The correct option is C $60\text{cm/s}$

Due to force $F$, the block starts moving in the direction of force and thus a restoring force is developed in the spring.
Let $x_{max}$ be the amplitude of the motion
The expression for restoring force is
$F=-k{x}_{max}$
$\Rightarrow x_{max}=\mid \frac{F}{k}\mid =\frac{10}{100}=0.1\text{m}=10\text{cm}$
Here, $k$ is the force constant
The angular frequency $\left(\omega \right)$ is given by
$\omega =\sqrt{\frac{k}{m}}\Rightarrow \omega =\sqrt{\frac{100}{1}}=\omega =10\text{rad/s}$
The expression for the speed of the block, when it is displaced by $x=8\text{cm}$ from its mean position is
$v=\omega \sqrt{x_{max}^2-x^2}$
$=10\sqrt{(10{)}^2-(8{)}^2}=60\text{cm/s}$
Thus, option (c) is the correct answer.