A two-pulley system allows a dockworker to apply a 1000 N force over a distance of 10 m, while lifting a crate by a vertical distance of 7.5 m. If the crate is raised at a constant speed and friction and resistance are negligible, what is the weight of the crate?

Question

Danny Avila

Physics

17 September, 05:17

A two-pulley system allows a dockworker to apply a 1000 N force over a distance of 10 m, while lifting a crate by a vertical distance of 7.5 m. If the crate is raised at a constant speed and friction and resistance are negligible, what is the weight of the crate?

+1

Answers (1)

Know the Answer?

Brennan Mcdonald · Answer 1

Weight of crate = 1000 N

Explanation:

Since the speed is constant, there is no acceleration in the system.

Also, since there is no friction and air resistance, the force needed to account for friction and air resistance is also 0.

So there is no need to take into account any horizontal movement of the crate, as no force is needed to continue that motion.

Vertical acceleration is also 0, and thus no extra force is needed to lift the object other than the weight of the object.

This is proved by newton's law: F = m*a

here a = 0 m/s^2

so F is also = 0 N

The only force required would be to counter the weight of the object:

F = Weight of crate

Substituting the value of force (F = 1000N) in this equation we get:

Weight of crate = 1000 N

Mass of crate = 1000 / 9.81 = 101.94 kg