A steel ball has a mass of 3.8 kg and rolls along a smooth, level surface at 56 m/s.

(a) Find its kinetic energy.
1 I got 404.32J bu it doset work
(b) At first, the ball was at rest on the surface. A constant force acted on it through a distance of 22 m to give it the speed of 56 m/s. What was the magnitude of the force?
2 N
-unsure how to calculate this(do you use your first answer?)

(a) KE = m•v²/2 = 3.8•(56)²/2 =5958.4 J.

(b) a =v²/2•s = (56)²/2•22 = 71.27 m/s².
F=m•a = 3.8•71.27 = 270.8 N

To calculate the kinetic energy of the steel ball, you can use the formula:

Kinetic energy (KE) = 1/2 * mass * velocity^2

(a) In this case, the mass of the steel ball is 3.8 kg and its velocity (speed) is 56 m/s. Plugging these values into the equation, we have:

KE = 1/2 * 3.8 kg * (56 m/s)^2
KE = 1/2 * 3.8 kg * 3136 m^2/s^2
KE ≈ 1/2 * 3.8 kg * 3136 J
KE ≈ 5964.8 J

So, the kinetic energy of the steel ball is approximately 5964.8 J, not 404.32 J.

For part (b), we need to calculate the magnitude of the force that acted on the ball through a distance of 22 m to give it the speed of 56 m/s.

To find the force, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.

Work (W) = Force (F) * Distance (d)

The work done on the ball is equal to the change in its kinetic energy:

W = KE final - KE initial

Since the ball started from rest initially, its initial kinetic energy is zero. Therefore, we have:

W = KE final - 0
W = KE final

Now, we know that the work done is equal to the force multiplied by the distance:

W = F * d

Therefore, we can write the equation as:

F * d = KE final

Plugging in the known values, we have:

F * 22 m = 5964.8 J

Now we can solve for the magnitude of the force (F):

F = 5964.8 J / 22 m
F ≈ 271.13 N

Therefore, the magnitude of the force acting on the steel ball was approximately 271.13 N, not 2 N.