A 190 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.160 m/s. How much work must be done on the hoop to stop it?
W =KE = m•v^2/2 + I•ω^2/2 =
= m•v^2/2 +(mR^2) •(v^2/R)/2= mv^2 =
=190•0.16^2 =4.86.4 J
To determine the amount of work required to stop the hoop, we need to use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The initial kinetic energy of the hoop is given by the formula:
KE_initial = 0.5 * m * v^2
where m is the mass of the hoop (190 kg) and v is its speed (0.160 m/s).
KE_initial = 0.5 * 190 kg * (0.160 m/s)^2
Now, we need to find the final kinetic energy of the hoop when it is at rest. Since the hoop is stopped, its final kinetic energy is zero.
KE_final = 0
Therefore, the work done on the hoop to stop it is equal to the change in kinetic energy:
Work = KE_final - KE_initial
Work = 0 - (0.5 * 190 kg * (0.160 m/s)^2)
Simplifying the equation, we have:
Work = -0.5 * 190 kg * (0.160 m/s)^2
Calculating the value yields the work required to stop the hoop. Keep in mind that the negative sign indicates that work is done against the hoop's motion.
Please note that the value of work will depend on the calculations, so make sure to substitute the given values correctly and perform the calculation accurately.