141k views
3 votes
A 1.8 kg block is pushed up against a spring whose spring constant is 650 N/m, compressing it 5.1 cm. When the block is released it moves horizontally without friction until it launches off the edge of a table 0.53 m above the ground. Find the horizontal distance covered by the block during its fall to the ground.

User Dokondr
by
7.9k points

1 Answer

6 votes

Final answer:

To find the horizontal distance covered by the block during its fall to the ground, we can calculate the time it takes for the block to reach the ground using the equation h = 0.5gt². We can then find the horizontal distance covered using the equation d = v * t, where v is the horizontal velocity and t is the time. Plugging in the given values, the horizontal distance covered is approximately 6.26 meters.

Step-by-step explanation:

To find the horizontal distance covered by the block during its fall to the ground, we can first determine the time it takes for the block to reach the ground.

Using the equation h = 0.5gt², where h is the height, g is the acceleration due to gravity (9.8 m/s²) and t is the time, we can solve for t. Plugging in the values, we get:

0.53 = 0.5 * 9.8 * t²

t² = 0.53 / (0.5 * 9.8)

t = sqrt(0.53 / (0.5 * 9.8))

t ≈ 0.34 s

Next, we can find the horizontal distance covered using the equation d = v * t, where d is the distance, v is the horizontal velocity, and t is the time.

The horizontal velocity can be found using the equation F = ma, where F is the force, m is the mass, and a is the acceleration. The force can be calculated using Hooke's Law, F = k * x, where k is the spring constant and x is the compression:

F = 650 * 0.051

F = 33.15 N

The acceleration can be found using a = F / m:

a = 33.15 / 1.8

a = 18.42 m/s²

Finally, plugging the values into the equation v = at:

v = 18.42 * 0.34

v ≈ 6.26 m/s

The horizontal distance covered by the block during its fall to the ground is 6.26 m.

User Gopesh
by
9.1k points