Question

A 2.6 kg block starts from rest at the top of a 30.9 —¦ incline and slides 2.02 m down the incline in 1.69 s. The acceleration of gravity is 9.8 m/s². What is the acceleration of the block?

Answer 1

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.

Step-by-step 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.