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Answer :
Final answer:
The acceleration of a 2.6 kg block sliding down a 30.9° incline is calculated using a kinematic equation and is found to be approximately 1.415 m/s².
Explanation:
The student's question concerns the acceleration of a block sliding down an incline. To find the block's acceleration, we can use the kinematic equation:
s = ut + \frac{1}{2}at^2,
where:
- s is the displacement (2.02 m),
- u is the initial velocity (0 m/s as the block starts from rest),
- t is the time (1.69 s), and
- a is the acceleration, which we want to find.
Plugging the values into the equation, we get:
2.02 m = 0 m/s (1.69 s) + \frac{1}{2}a(1.69 s)^2.
We can solve for 'a' to get the acceleration of the block. The calculation gives us:
a = \frac{2(2.02 m)}{(1.69 s)^2} = \frac{4.04 m}{(2.8541 s^2)} ≈ 1.415 m/s^2.
So, the acceleration of the block is approximately 1.415 m/s².
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