# A 2.5 kg box is held released from rest 1.5 meters above the ground and slides down a...

## Question:

A 2.5 kg box is held released from rest 1.5 meters above the ground and slides down a frictionless ramp. It slides across a floor that is frictionless, except for a small section 0.5 meters wide that has a coefficient of kinetic friction of 0.4. At the left end, is a spring with spring constant 250 N/m. The box compresses the spring and is accelerated back to the right. What is the maximum height to which the box returns on the ramp?

## Conservation Of Energy :

In the conservation of energy principle, the total amount of energy contained within the system remain conserved during the conversion process that means the energy cannot be created and cannot be destroyed and will be converted from one form into another for example potential energy contained within the system can be converted into the kinetic energy when the object is allowed to fall.

Given

Mass of the box is {eq}m= 2.5\ kg {/eq}

Height of the object from the ground is {eq}h= 1.5\ m {/eq}

Coefficient of kinetic friction is {eq}\mu_k= 0.4 {/eq}

Spring constant of the spring is {eq}k= 250\ N/m {/eq}

Now from the energy conservation principle Potential energy at the top will equal to the frictional energy loss and compression in the spring :

{eq}P.E=F.E+F.E\\ mgh=\mu(mg)\times d+\frac{1}{2}kx^2\\ 2.5\times 9.81\times 1.5=0.4\times (2.5\times 9.81)\times 0.5+\frac{1}{2}(250)\times x^2\\ 36.78=4.905+125\ x^2\\ x=0.504\ m {/eq}

Thus, the compression of the spring is 0.504 m

{eq}\underline{\textrm{Now for the height that the box returns:}} {/eq}

{eq}\frac{1}{2}kx^2-\mu(mg)\times d=mgH\\ \frac{1}{2}250\times (0.504)^2-0.4\times (2.5\times 9.81)\times 0.5=2.5\times 9.81\times H\\ H= 1.09\ m {/eq}

Thus, the maxium height to which the box return on the ramp is 1.09 m 