A block is released from rest at the top of a frictionless inclined plane 16 m long. It reaches the bottom 4.2 s later. A second block is projected up the plane from the bottom at the instant the first block is released in such a way that it returns to the bottom simultaneously with the first block. (a) Find the acceleration of each block on the incline. (b) What is the initial velocity of the second block? (c) How far up the incline does it travel? You can assume that both blocks experience the same acceleration.

This problem involves two blocks moving along a frictionless inclined plane. The first block is released from rest at the top of the incline, and we are given its travel time to the bottom. The second block is projected upward from the bottom at the same time the first block is released, and it returns to the bottom at the same time as the first block. Our goal is to find the acceleration of both blocks, the initial velocity of the second block, and the distance it travels up the incline.

Given Data

• Length of the incline (L): 16 m
• Time for the first block to reach the bottom (t): 4.2 s
• Acceleration due to gravity (g): 9.8 m/s² (assumed for Earth)

(a) Find the acceleration of each block on the incline

To find the acceleration of the blocks on the incline, we will use the kinematic equation that relates distance, time, and acceleration:

L = ½ a t²

Where:

• L is the distance traveled along the incline (16 m),
• a is the acceleration on the incline (which we need to find),
• t is the time taken (4.2 s).

Rearranging the equation to solve for acceleration a:

a = 2L / t²

Substituting the known values:

a = 2 × 16 m / (4.2 s)²

a ≈ 2 × 16 m / 17.64 s²

a ≈ 32 m / 17.64 s² ≈ 1.81 m/s²

So, the acceleration of each block on the incline is approximately 1.81 m/s².

(b) What is the initial velocity of the second block?

For the second block, we know that it is projected upward from the bottom of the incline at the same time the first block is released, and it returns to the bottom at the same time as the first block. This means that the total time the second block travels (up and down) is also 4.2 s.

We can use the kinematic equation for the second block moving upward:

0 = v₀² - 2a d Where:

• v₀ is the initial velocity of the second block (which we need to find),
• a is the acceleration (1.81 m/s²),
• d is the distance traveled upward (which is the distance the block moves up the incline before stopping, i.e., the maximum height reached by the second block).

We know that the total time for the second block to travel upward and then return to the bottom is 4.2 s. Since the time to go up equals the time to come down, the time for the second block to travel upward is:

t_up = 4.2 s / 2 = 2.1 s

Now, we can use the kinematic equation to find the distance traveled upward:

d = v₀ × t_up - ½ a t_up²

Substituting the known values:

d = v₀ × 2.1 s - ½ × 1.81 m/s² × (2.1 s)²

d = v₀ × 2.1 s - ½ × 1.81 m/s² × 4.41 s²

d = v₀ × 2.1 s - 3.99 m

But we also know that the distance traveled upward is equal to the length of the incline (16 m) since the second block goes up and then comes back down the same distance. Therefore, we have:

16 m = v₀ × 2.1 s - 3.99 m

Solving for v₀:

v₀ × 2.1 s = 16 m + 3.99 m = 19.99 m

v₀ = 19.99 m / 2.1 s ≈ 9.52 m/s

Thus, the initial velocity of the second block is approximately 9.52 m/s.

(c) How far up the incline does the second block travel?

We already calculated that the second block travels a distance d upward before it stops, which is the maximum height the block reaches. From part (b), we know that the distance traveled upward is:

d = 16 m

Therefore, the second block travels 16 meters up the incline before stopping and then returns to the bottom.

Summary of Results

• Acceleration of each block: 1.81 m/s²
• Initial velocity of the second block: 9.52 m/s
• Distance traveled by the second block up the incline: 16 m