# You compress a spring with k = 3000 N/m a distance of 0.2 m with a block of mass 4 kg. You then...

## Question:

You compress a spring with k = 3000 N/m a distance of 0.2 m with a block of mass 4 kg. You then release the block. It slides along a frictionless surface and then goes up an incline of 35{eq}^{\circ} {/eq}. What is the velocity of the block after it leaves the spring but before it gets to the incline?

How far up the incline (along the hypotenuse of the triangle) does the block travel?

## Kinetic energy:

The term kinetic energy is defined as considered when the body is in motion for producing some work done. The kinetic energy is used to accelerate the body with a velocity. The standard unit of the kinetic energy is Newton meter.

Given data:

• The spring stiffness is {eq}k = 3000\,{\rm{N/m}} {/eq}
• The block of mass is {eq}m = 4\,{\rm{kg}} {/eq}
• The distance is {eq}d = 0.2\,{\rm{m}} {/eq}
• The inclination angle is {eq}\theta = 35^\circ {/eq}

The expression for the velocity of the block is by using the energy conservation principle is

{eq}w = k.e {/eq}

• Here {eq}w = \dfrac{1}{2}k{d^2} {/eq}is the work done by the spring.
• Here {eq}k.e = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.

Substituting the values in the above equation as,

{eq}\begin{align*} w = k.e\\ \dfrac{1}{2}k{d^2} &= \dfrac{1}{2}m{v^2}\\ \left( {3000} \right){\left( {0.2} \right)^2} &= \left( 4 \right){v^2}\\ v &= 5.48\,{\rm{m/s}} \end{align*} {/eq}

Thus the velocity of the block is {eq}v = 5.48\,{\rm{m/s}} {/eq}

The expression for the block travels on the incline is by using the energy conservation principle is

{eq}k.e = p.e {/eq}

• Here {eq}k.e = \dfrac{1}{2}m{v^2} {/eq} is the kinetic energy.
• Here {eq}p.e = mgh {/eq} is the potential energy of the system.

Substituting the values in the above equation as,

{eq}\begin{align*} k.e &= p.e\\ \dfrac{1}{2}m{v^2} &= mgh\\ {v^2} &= gh \end{align*} {/eq}

• Here {eq}h = L\sin \theta {/eq}

Substitute the values in the above equation as,

{eq}\begin{align*} {v^2} &= gh\\ {v^2} &= g \times L\sin \theta \\ {\left( {5.48} \right)^2} &= 9.8 \times L\sin 35^\circ \\ L &= 2.67\,{\rm{m}} \end{align*} {/eq}

Thus the block travels is {eq}L = 2.67\,{\rm{m}} {/eq}