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 coefficient of kinetic friction between block and tabletop is $0.20$.

• A. $7.38m/s$
• B. $2.38m/s$
• C. $12.38m/s$
• D. $8.38m/s$

Answer 1

The correct option is B $2.38m/s$

Frictional force$=\mu N=0.2\times 15=3N$

Applying Work Energy Theorem we get

$W_{A"}=△K-\frac{1}{2}\left(19.6\right){\left(0.3\right)}^2-(3\times 0.3)+(20\times 0.3)=\frac{1}{2}\left(1.5\right)v^2-0.88-0.9+6=\frac{1}{2}\times \left(0.5\right)v^{2}v=4m/s$