A 100 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.230 m/s. How much work must be done on the hoop to stop it?
m*v^2
To find the work done on the hoop to stop it, we need to consider the kinetic energy of the hoop.
The kinetic energy of an object is given by the equation:
KE = (1/2) * m * v^2
Where:
KE is the kinetic energy
m is the mass of the object
v is the velocity of the object
In this case, the mass of the hoop is given as 100 kg and the velocity is given as 0.230 m/s.
Substituting these values into the equation:
KE = (1/2) * 100 kg * (0.230 m/s)^2
Now we can calculate the kinetic energy:
KE = 0.5 * 100 kg * 0.053 m^2/s^2
KE = 2.65 J
The work done on the hoop is equal to the change in kinetic energy. Since we want to stop the hoop completely, the final kinetic energy is zero.
So, the work done to stop the hoop is:
Work = KE_final - KE_initial
Work = 0 - 2.65 J
Work = -2.65 J
Therefore, to stop the hoop, 2.65 Joules of work must be done on it.