block a of mass 4kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring constant 650n/m. the other end of the spring is attacked to a wall. the block is pushed toward the wall until the spring has been compressed a distance x. the block is released and follows the trajectory falling 0.80m vertically and stricking a target on the floor that is a horizontal distance of 1.3 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 s of the block as it leaves the table. C)calculate the distance x the spring was ccompressed.

A) 0.80 m/s^2 * (1.3 m + 0.80 m) / 2 = 1.04 s

B) 4 kg * 9.8 m/s^2 * (1.3 m + 0.80 m) / 2 = 39.2 m/s

C) x = (4 kg * 9.8 m/s^2 * (1.3 m + 0.80 m)^2) / (2 * 650 N/m) = 0.45 m

A) To calculate the time elapsed from when block A leaves the table to when it strikes the floor, we can use the equation for vertical displacement:

h = 0.5 * g * t^2

In this equation, h represents the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time elapsed. Given that the block fell a distance of 0.80 m, we can rearrange the equation to solve for t:

t^2 = (2 * h) / g

t^2 = (2 * 0.80 m) / 9.8 m/s^2
t^2 = 0.1633

Taking the square root of both sides:

t = sqrt(0.1633) s
t ≈ 0.404 s

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

B) To calculate the speed of the block as it leaves the table, we can use the principle of conservation of mechanical energy. Initially, the block has only potential energy stored in the compressed spring, which is then converted entirely into kinetic energy as it leaves the table. The equation for kinetic energy is:

KE = 0.5 * m * v^2

In this equation, KE represents kinetic energy, m is the mass of the block (4 kg), and v is the speed of the block as it leaves the table. The potential energy stored in the spring is given by:

PE = 0.5 * k * x^2

Where PE represents potential energy, k is the spring constant (650 N/m), and x is the distance the spring was compressed.

Since the potential energy is entirely converted into kinetic energy, we can set these two equations equal to each other and solve for v:

0.5 * k * x^2 = 0.5 * m * v^2

(650 N/m) * x^2 = (4 kg) * v^2

Simplifying:

v^2 = (650 N/m) * x^2 / (4 kg)

v^2 = (162.5 N/m) * x^2 / kg

v = sqrt((162.5 N/m) * x^2 / kg)

C) To calculate the distance x the spring was compressed, we can use the equation for potential energy stored in the spring:

PE = 0.5 * k * x^2

In this equation, PE represents potential energy, k is the spring constant (650 N/m), and x is the distance the spring was compressed. We can rearrange the equation to solve for x:

x^2 = 2 * PE / k

x^2 = 2 * (m * g * h) / k

x^2 = 2 * (4 kg) * (9.8 m/s^2) * (0.80 m) / (650 N/m)

x^2 ≈ 0.096 m^2

Taking the square root of both sides:

x ≈ sqrt(0.096) m
x ≈ 0.31 m

Therefore, the distance x the spring was compressed is approximately 0.31 meters.

To solve this problem, we need to analyze the motion of the block before it leaves the table and after it leaves the table.

A) To calculate the time elapsed from the instant the block leaves the table to the instant it strikes the floor, we need to find the time it takes for the block to reach the floor after leaving the table.

First, we can calculate the initial velocity of the block as it leaves the table. Since the table is frictionless, the only force acting on the block at that moment is the force provided by the spring.

The force provided by the spring can be calculated using Hooke's Law:

F = k * x

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

Using Newton's second law (F = ma), we can equate the force provided by the spring to the acceleration of the block:

k * x = m * a

Simplifying this equation, we get:

a = (k * x) / m

Now, let's analyze the motion of the block after it leaves the table. The vertical motion can be analyzed independently, as there is no horizontal force acting on the block. The time it takes for the block to fall vertically can be found using the equation:

y = (1/2) * g * t^2

where y is the vertical distance (0.80 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Rearranging the equation, we get:

t^2 = (2 * y) / g

t = sqrt((2 * y) / g)

Next, we need to find the horizontal distance the block travels during this time. The horizontal velocity of the block remains constant (assuming no air resistance), so we can use the equation:

x = v * t

where x is the horizontal distance (1.3 m), v is the velocity, and t is the time.

From the previous step, we found the time (t) it takes for the block to fall vertically. Now, we need to calculate the horizontal velocity (v). To do this, we need to find the horizontal acceleration of the block using the value of the force provided by the spring.

The horizontal acceleration of the block is given by:

a_horizontal = (k * x) / m

Now, we can use this acceleration to find the horizontal velocity:

v = a_horizontal * t

B) To calculate the speed of the block as it leaves the table, we need to find the magnitude of the velocity. Since the block moves horizontally, the magnitude of the velocity will be the horizontal velocity we calculated in part A.

C) To calculate the distance x the spring was compressed, we need to rearrange the equation from part A:

x = (m * a) / k

Substituting the known values, we can calculate the distance x.

Please provide the values of the mass (m) and distance (x) for a more specific calculation.