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?

1 Answer
Jun 12, 2015

Answer:

#v=4.80m/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_0t+1/2at^2#

and

#v=v_0+at#.

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+2aDeltas# in which #Deltas# is the space run.

So:

#v=sqrt(v_0^2+2aDeltas)=sqrt(0^2+2*5.48009*2.1)=4.80m/s#.