Question

The spring has a force constant $24N/m$ . The mass of the block attached = $4kg$ . Initially the block is at rest & spring is unstretched. If a constant horizontal force of $10N$ is applied, what is the speed of the block when it has been moved to a distance of $5m$

Answer 1

The equation of motion is given as,

$F-kx=ma$

$10-24\times 5=4\times a$

$a=-27.5m/s^2$

The speed is given as,

$v^2=u^2+2ax$

$v^2={\left(0\right)}^2+2\times 27.5\times 5$

$v=16.58m/s$

Thus, the speed of the block is $16.58m/s$.