At t=1.0s, a 0.49 kg object is falling with a speed of 6.5 m/s. At t=2.0s, it has a kinetic energy of 38 J.

What is the kinetic energy of the object at t=1.0s?
What is the speed of the object at t=2.0s?
How much work was done on the object between t=1.0s and t=2.0s?

To answer these questions, we need to understand the concepts of kinetic energy and work.

1. What is the kinetic energy of the object at t=1.0s?

The formula for kinetic energy is: KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass of the object, and v is the velocity/speed of the object.

Given:
m = 0.49 kg
v = 6.5 m/s

Substituting these values into the formula, we can calculate the kinetic energy:
KE = (1/2) * 0.49 kg * (6.5 m/s)^2

Calculating this, we find:
KE ≈ 10.05825 J

Therefore, the kinetic energy of the object at t=1.0s is approximately 10.05825 J.

2. What is the speed of the object at t=2.0s?

Unfortunately, we don't have enough information to directly calculate the speed at t=2.0s because the problem did not provide any data about the acceleration or forces acting on the object. We only know the kinetic energy at that particular time.

3. How much work was done on the object between t=1.0s and t=2.0s?

The work done on an object can be calculated using the formula: Work = ΔKE
where ΔKE represents the change in kinetic energy.

Given:
Initial kinetic energy (at t=1.0s) = 10.05825 J
Final kinetic energy (at t=2.0s) = 38 J

Substituting these values into the formula, we can calculate the work done:
Work = 38 J - 10.05825 J

Calculating this, we find:
Work ≈ 27.94175 J

Therefore, approximately 27.94175 J of work was done on the object between t=1.0s and t=2.0s.