# A 60 N crate starting at rest slides down a rough 5.0 m ramp, inclined at 25 degrees with the...

## Question:

A 60 N crate starting at rest slides down a rough 5.0 m ramp, inclined at 25 degrees with the horizontal. 20 J of energy is lost due to friction. What will be the speed of the crate at the bottom of the incline?

## Law of conservation of energy

This states that without the presence of an external force, the initial energy of the system is equal to its final energy hence energy is conserved .Here in this problem, there exist an external force which is friction

Given:

• Weight of the crate {eq}w = 60 \rm\ N{/eq}
• Length of the ramp {eq}l = 5 \rm\ m{/eq}
• Angle {eq}\theta = 25 ^{\circ}{/eq}
• Energy lost {eq}E = 20 \rm\ J{/eq}

The height of the ramp is

\begin{align} sin\theta &= \frac{y}{l}\\ y &= lsin\theta\\ y &= (5 \rm\ m)sin (25)\\ y &= 2.11 \rm\ m \end{align}

The initial energy of the system is

\begin{align} E_i &= KE_i + PE_i\\ E_i &= 0 + mgy\\ E_i &= (60 \rm\ N)(2.11 \rm\ m)\\ E_i &= 126.6 \rm\ J \end{align}

Now, we will compute for velocity. Note that there is an energy lost

\begin{align} E_f&= KE_f + PE_f\\ 126.6 \rm\ J - 20 \rm\ J &= \frac{1}{2}mv^2 + mgy\\ 106.6 \rm\ J &= \frac{1}{2}(\frac{60 \rm\ N}{9.8 \rm\ m/s^2})v^2 + 0\\ v &= \sqrt{\frac{213.2 \rm\ J}{6.12 \rm\ kg}}\\ v &= \boxed{5.9 \rm\ m/s} \end{align}