Question

A ball of mass 2kg is dropped from a height of 5.0 m above the floor. Find the speed of the ball as it strikes the floor.

Answer 1

Using the conservation of energy, the speed of the 2kg ball before striking the floor is approximately 9.9 m/s after being dropped from a height of 5.0 m.

Step-by-step explanation:

The student asked about the speed of a 2kg ball after being dropped from a height of 5.0 m and striking the floor. To solve this problem, we use the principles of physics related to gravitational potential energy and kinetic energy. Assuming air resistance is negligible, all potential energy of the ball at the height of 5.0 m is converted to kinetic energy just before it strikes the floor.

Using the equation of conservation of energy, we have:
Potential Energy (PE) = Kinetic Energy (KE)
mgh = (1/2)mv²

where m is mass, g is the acceleration due to gravity (9.8 m/s²), h is height, and v is final velocity.

We can find the velocity as follows:

2 * 9.8 * 5 = (1/2) * 2 * v²
98 = v²
v ≈ 9.9 m/s (since v is positive for a falling object)

Thus, the speed of the ball as it strikes the floor is approximately 9.9 meters per second.

Answer 2

The speed of the ball as it strikes the floor is approximately 9.9 m/s.

Step-by-step explanation:

Calculate the potential energy:

Potential energy (PE) = mass (m) x gravity (g) x height (h)

PE = 2 kg x 9.81 m/s² x 5.0 m

PE = 98.1 Joules

Equate potential energy to kinetic energy:

At the point of impact, all the potential energy is converted into kinetic energy (KE).

KE = 1/2 x mass (m) x velocity (v)²

Since the ball starts from rest, its initial velocity (u) is 0.

Solve for the velocity (v):

98.1 Joules = 1/2 x 2 kg x v²

v² = 98.1 Joules / (1/2 x 2 kg)

v² = 98.1 Joules / 1 kg

v² = 98.1 m²/s²

v = √98.1 m²/s²

v ≈ 9.9 m/s

Therefore, the speed of the ball as it strikes the floor is approximately 9.9 m/s.