A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth?

Explanation

Linear velocity v=72 km/h = `(72 xx 1000)/(60 xx 60) = 20` m/s
Initial angular velocity of the wheel `=omega_(0) = v/r = 20/(0.25) = 80` rad/s
Final angular velocity =`omega =0`
Time =t= 16s
Angular acceleration = `alpha` = ?
`alpha = (omega-omega_(0))/t = (0-80)/16 = -5 "rad"//s^(2)`
Average torque `=tau = lalpha = 5 xx 5 = 25` Nm