Question

Problem 5:
The block m of 2 Kg is on a plan inclined at an angle θ = 35.0 ° to the horizontal, with a friction
coefficient = 0.22.
- If the block starts from rest at the top of the incline which is 10 m above the horizontal
base, what will be the velocity of the block when it reaches the middle of the incline?

Answer 1

As the block starts from rest, it has only potential energy at the height.

• $\sf{K_i=0}$

• $\sf{U_i=mgh}$

As the block reaches the middle of the incline, now it's height is $\sf{\dfrac{h}{2}}$ and speed is $\sf{v.}$ So,

• $\sf{K_f=\dfrac{1}{2}\,mv^2}$

• $\sf{U_f=\dfrac{1}{2}\,mgh}$

By energy conservation,

$\sf{\longrightarrow K_i+U_i=K_f+U_f}$

$\sf{\longrightarrow 0+mgh=\dfrac{1}{2}\,mv^2+\dfrac{1}{2}\,mgh}$

$\sf{\longrightarrow \dfrac{1}{2}\,mgh=\dfrac{1}{2}\,mv^2}$

$\sf{\longrightarrow v=\sqrt{gh}}$

In this question, $\sf{h=10\ m.}$ Taking $\sf{g=10\ m\,s^{-2},}$

$\sf{\longrightarrow v=\sqrt{10\times10}}$

$\sf{\longrightarrow\underline{\underline{v=10\ m\,s^{-1}}}}$