A 2.01 kg block starts from rest at the top of a 25.8degree incline and accelerates uniformly down the incline, moving 2.13m in 2.98 seconds.
What is the magnitude of the acceleration of the block?
distance=1/2 a t^2 solve for a
To find the magnitude of the acceleration of the block, we can use the kinematic equation:
\[s = ut + \frac{1}{2}at^2\]
where:
- \(s\) represents the displacement of the block (2.13 m in this case)
- \(u\) represents the initial velocity of the block (0 since it starts from rest)
- \(t\) represents the time taken (2.98 seconds in this case)
- \(a\) represents the acceleration of the block
Rearranging the equation to solve for acceleration (\(a\)) gives us:
\[a = \frac{2(s - ut)}{t^2}\]
Now, let's substitute the given values into the equation:
\[a = \frac{2(2.13 - 0\cdot2.98)}{(2.98)^2}\]
Simplifying further:
\[a = \frac{2\cdot2.13}{(2.98)^2}\]
Evaluating this expression:
\[a = \frac{4.26}{8.88}\]
Finally, calculating the magnitude of the acceleration (the absolute value):
\[a = 0.479 \, \text{m/s}^2\]
So, the magnitude of the acceleration of the block is 0.479 m/s².