A 2.67 kg block starts from rest at the top of a
27.6� incline and accelerates uniformly down
the incline, moving 2.89 m in 2.45 s.
The acceleration of gravity is 9.81 m/s2 .
a) What is the magnitude of the accelera-
tion of the block?
Answer in units of m/s2
To find the magnitude of the acceleration of the block, we can use the kinematic equation that relates displacement, initial velocity, acceleration, and time:
\(s = ut + \frac{1}{2}a t^2\)
In this equation, s represents the displacement, u represents the initial velocity, a represents the acceleration, and t represents the time.
Since the block starts from rest (u = 0), the equation simplifies to:
\(s = \frac{1}{2} a t^2\)
Rearranging the equation to solve for acceleration (a) gives:
\(a = \frac{2s}{t^2}\)
Now we can plug in the given values into this equation:
s = 2.89 m
t = 2.45 s
\(a = \frac{2 \times 2.89 \,m}{(2.45 \,s)^2}\)
Calculating this expression gives:
\(a \approx 1.203 \, \text{m/s}^2\)
Therefore, the magnitude of the acceleration of the block is approximately 1.203 m/s².