The potential energy of the box when it gets to the top is
(mass) (gravity) (height)
= (7 kg) (9.8 m/s²) (5 m)
= 343 joules.
That's the work done against the force of gravity. Any
additional work is done against the force of friction.
Answer:
20 cm to the right of the center or 20+50 = 70 cm from the left side.
Explanation:
The length of meter stick is 1 m = 100 cm
Balance point on 50 cm
From the center the 20 N weight is 50-20 = 30 cm
Torque is obtained when force is multiplied with the distance
As the force is conserved we have

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.
Based on the situation above the the work done was 400 Joules. <span>Q = FS cos(theta) is the so-called work function. It's important to learn the work physics; you'll see it over and over in science/physics class. Theta is the angle between the force vector F and the distance vector S. In your problem we assume theta = 0, the two vectors were assumed aligned.</span>