A steel ball has a mass of 4.0kg and rolls along a smooth, level surface at 62m/s.
A) Find its kinetic energy(i got this and its 7688 joules)
B)?????-At first, the ball was at rest on the surface. A constant force acted on it through a distance of 22 meters to give it the speed of 62m/s. What was the magnitude of the force??
I really don't understand part b. please help in detail! Thxs PS- i need to use energy work formuals. no kinematics formuals
Ignoring ball rotation and friction, the work done is F * X and equals the final kinetic energy, (1/2) MV^2
Therefore F = (1/2)MV^2/X
X= 22m
When rotational kinetic energy is considered, (3/5)MV^2 replaces (1/2)MV^2, but they probably do not expect you to know that.
so whats the final answer??
To find the magnitude of the force that acted on the steel ball, we can use the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy. In this case, the work done on the ball is the product of the force applied and the distance over which the force is applied.
The work done on the ball is given by the formula:
Work = Force × Distance
We know the distance over which the force is applied is 22 meters. We need to find the force that was applied.
Since the ball starts from rest and reaches a final velocity of 62 m/s, the change in kinetic energy is equal to the final kinetic energy (which we already found to be 7688 Joules) minus the initial kinetic energy (which is zero since it starts from rest).
Change in Kinetic Energy = Final Kinetic Energy - Initial Kinetic Energy
Change in Kinetic Energy = 7688 J - 0 J
Change in Kinetic Energy = 7688 J
According to the work-energy principle, the change in kinetic energy is also equal to the work done on the ball. So we can rewrite the equation as:
Work Done = Change in Kinetic Energy
Plugging in the known values:
Force × Distance = 7688 J
Since the distance is given as 22 meters, we can rearrange the equation to solve for the force:
Force = Change in Kinetic Energy / Distance
Force = 7688 J / 22 m
Force = 349.45 N
So the magnitude of the force that acted on the steel ball is approximately 349.45 Newtons.