We have that the Force is mathematically given as
F=170.833N
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Force </h3>
Question Parameters:
- The torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 41 N · m,
- the end of a 24 cm-long wrench to <em>loosen </em>the nut
Generally the equation for the Force is mathematically given as

Therefore
F=41/0.24
F=170.833N
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<span>Total KE = KE (rotational) + KE (translational)
Moment of inertia of sphere is I = (2/5)mr^2
So KE (rotational) = (1/2) x I x w^2 = (1/2) x (2/5)mr^2 x w^2 = (1/5) x m x r^2 x w^2
KE (translational) = (1/2) x m x v^2 = (1/2) x m x (rw)^2 = (1/2) x m x r^2 x w^2
Hence KE = (1/5) x m x r^2 x w^2 + (1/2) x m x r^2 x w^2 = m x r^2 x w^2 ((1/5) + (1/2))
KE = (7/10) m x r^2 x w^2
Calculating the fraction of rotational kinetic energy to total kinetic energy,
= rotational kinetic energy / total kinetic energy
= (1/5) x m x r^2 x w^2 / (7/10) m x r^2 x w^2 = (1/5) / (7/10) = 2 / 7
The answer is 2 / 7</span>