A 65 kg crate slides on a rough plane inclined upward at anangle of 28 degrees.

Question

A 65 kg crate slides on a rough plane inclined upward at anangle of 28 degrees. The crate is hauled up the plane mylightweight rope. A worker standing below the pulley pullsvertically downwars on the rope. If the 55 kg worker hangs his other entire weight fom the rope, it is barely sufficient to move thecrate up the incline at a constant velocit of 1.2 m/s. (a) What is the magnitude of the sliding friction force on thebox? (b) What is the coefficient of kinetic friction between thebox and the incline? A 65 kg crate slides on a rough plane inclined upward at anangle of 28 degrees. The crate is hauled up the plane mylightweight rope. A worker standing below the pulley pullsvertically downwars on the rope. If the 55 kg worker hangs his other entire weight fom the rope, it is barely sufficient to move thecrate up the incline at a constant velocit of 1.2 m/s. (a) What is the magnitude of the sliding friction force on thebox? (b) What is the coefficient of kinetic friction between thebox and the incline?

Explanation / Answer

The tension in the rope is T=mg=55*9.8=539N T=mgsin+fr 539=65*9.8*sin28+fr fr=239.946614N fr=kmgcos28 k=239.946614/65*9.8cos28 k=0.4266 Hence we get by it.

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