Question

A $1.5$ kg block at rest on a tabletop is attached to a horizontal spring having a spring constant of $19.6$ N/m. The spring is initially unstretched. A constant $20.0$ N horizontal force is applied to the object causing the spring to stretch.Determine the speed of the block after it has moved $0.30$ m from equilibrium if the surface between the block and the tabletop is frictionless.

• A. $2.61m/s$
• B. $3.61m/s$
• C. $7.61m/s$
• D. $8.1m/s$

Answer 1

The correct option is B $3.61m/s$

This system will exhibits S.H.M with angular frequency $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{19.6}{1.5}}$

$k$= spring constant

$m$= mass of the body

The string stretched by the maximum A restoring force at maximum stretch= force acting on body

$\Rightarrow kA=F$

$\Rightarrow A=\frac{F}{k}=\frac{20}{19.6}$

$F=20N$

$K=19.6N/m$

Now if x is displacement from mean position, the velocity is given by:

$v=\omega =\sqrt{(A^2-x^2)}$

$v=\omega =\sqrt{\frac{19.6}{1.5}({\frac{20}{19.6}}^2-{0.3}^2)}$

$=3.523m/s$