Block A of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass

and spring constant 650 N m. The other end of the spring is attached to a wall. The block is pushed toward the
wall until the spring has been compressed a distance x, as shown above. The block is released and follows the
trajectory shown, falling 0.80 m vertically and striking a target on the floor that is a horizontal distance of 1.2 m
from the edge of the table. Air resistance is negligible.

(a) Calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor.

(b) Calculate the speed of the block as it leaves the table.

(c) Calculate the distance x the spring was compressed.

Block B, also of mass 4.0 kg, is now placed at the edge of the table. The spring is again compressed a distance x,
and block A is released. As it nears the end of the table, it instantaneously collides with and sticks to block B.
The blocks follow the trajectory shown in the figure below and strike the floor at a horizontal distance d from the
edge of the table.

To solve this problem, we need to break it down into smaller parts and use the relevant equations of motion and energy conservation principles.

(a) Calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor:

First, we need to calculate the time it takes for block A to fall vertically and strike the floor. We can use the equation for vertical motion with constant acceleration:

h = (1/2)gt^2,

where h is the vertical distance (0.80 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Rearranging the equation for time:

t = sqrt(2h/g).

Now we can calculate the time it takes for block A to move horizontally from the edge of the table to the floor. This time will be the same as the time calculated above because the motion is independent of the horizontal distance. Let's call this time T.

Therefore, the total elapsed time is:

Total time = 2T.

(b) Calculate the speed of the block as it leaves the table:

To determine the speed of block A as it leaves the table, we can use the principle of conservation of mechanical energy.

The initial potential energy stored in the compressed spring is converted entirely into the kinetic energy of the block as it leaves the table.

The potential energy stored in the spring is given by:

PE = (1/2)kx^2,

where k is the spring constant (650 N/m) and x is the distance the spring was compressed.

The kinetic energy of the block as it leaves the table is given by:

KE = (1/2)mv^2,

where m is the mass of block A (4.0 kg) and v is its velocity.

Since energy is conserved, we can equate the potential energy to the kinetic energy:

(1/2)kx^2 = (1/2)mv^2.

Simplifying and solving for v:

v = sqrt((k/m)x^2).

(c) Calculate the distance x the spring was compressed:

To find the distance x, we need to use the equation for potential energy stored in the spring:

PE = (1/2)kx^2.

Rearranging the equation and plugging in the known values:

x = sqrt((2PE)/k).

In this case, PE is the initial potential energy stored in the compressed spring, which will be converted into kinetic energy as the block leaves the table.

To solve this problem, we can use energy conservation and kinematic equations. Let's break it down step by step:

Step 1: Calculate the time elapsed from the instant block A leaves the table to the instant it strikes the floor.

We can use the equation for vertical displacement:
y = vi*t + (1/2) * a * t^2

Given:
Initial vertical velocity (vi) = 0 m/s
Vertical displacement (y) = -0.80 m (negative since it's downward)
Acceleration (a) = 9.8 m/s^2 (acceleration due to gravity)

Plugging in the values:
-0.80 = 0 * t + (1/2) * 9.8 * t^2

Simplifying the equation:
4.9 * t^2 = 0.80

Solving for t:
t^2 = 0.80 / 4.9
t^2 = 0.163265
t ≈ √0.163265
t ≈ 0.404 s

So, the time elapsed from the instant block A leaves the table to the instant it strikes the floor is approximately 0.404 seconds.

Step 2: Calculate the speed of the block as it leaves the table.

We can use the equation for energy conservation:
Potential energy (PE) = Kinetic energy (KE)

The potential energy stored in the compressed spring is given by:
PE = (1/2) * k * x^2

The kinetic energy of the block as it leaves the table is given by:
KE = (1/2) * m * v^2

Where:
k = spring constant = 650 N/m
x = distance the spring was compressed
m = mass of the block = 4.0 kg
v = velocity of the block as it leaves the table (what we need to find)

Since the block starts from rest on the table, it has no initial kinetic energy.

Therefore, we can write:
(1/2) * k * x^2 = (1/2) * m * v^2

Simplifying the equation:
650 * x^2 = 4 * v^2

Taking the square root of both sides:
v = √(650/4) * x
v = √(162.5) * x
v ≈ 12.74 * x

So, the speed of the block as it leaves the table is approximately 12.74 times the distance the spring was compressed.

Step 3: Calculate the distance x the spring was compressed.

Given:
Vertical distance from the table to the floor (y) = 0.80 m
Horizontal distance from the edge of the table to the target (d) = 1.2 m

Since there is no horizontal acceleration, the time it takes for the block to travel horizontally is the same as the time it takes to fall vertically.

Using the formula for horizontal distance:
d = v * t

Substituting the value of t (0.404 s) from step 1:
1.2 = v * 0.404

Rearranging the equation:
v = 1.2 / 0.404
v ≈ 2.972 m/s

Now, we can use the equation for horizontal distance:
d = v * t

Substituting the value of v (2.972 m/s) and t (0.404 s):
1.2 = 2.972 * 0.404

Rearranging the equation:
d ≈ 1.2 / 2.972
d ≈ 0.403 m

So, the distance x the spring was compressed is approximately 0.403 meters.

yolo