# A 2.67 kg block slides down a frictionless plane, from rest, with an acceleration of 5.48009 m/s^2. What's the block's speed (in m/s) after travelling 2.1 m along the incline?

Jun 12, 2015

$v = 4.80 \frac{m}{s}$.

#### Explanation:

The facts that the block is going down a plane is not so important. The important thing is that the plane is frictionless.

The two laws of the accelerated motion are:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} a {t}^{2}$

and

$v = {v}_{0} + a t$.

But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.

${v}^{2} = {v}_{0}^{2} + 2 a \Delta s$ in which $\Delta s$ is the space run.

So:

$v = \sqrt{{v}_{0}^{2} + 2 a \Delta s} = \sqrt{{0}^{2} + 2 \cdot 5.48009 \cdot 2.1} = 4.80 \frac{m}{s}$.