A massless string runs around two massless, frictionless pulleys. An object with mass m = 20 kg hangs from one pulley. A force F is exerted on the free end of the string to lift the object at a constant speed by a distance of 2.3 m. What is the work done on the object by the force F? Hint: Use the lower pulley and the attached object as the system and note that the force F is the tension in the string.

work done on gravity= 20*g*2.3m

This has to be the work done by F also.

To find the work done on the object by the force F, we can use the formula:

Work = Force * Distance * Cos(θ)

In this case, the force F is the tension in the string which is responsible for lifting the object. Since the object is being lifted at a constant speed, we know that the net force acting on the object is zero. This means the force F is equal to the weight of the object:

Force F = m * g

where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s²).

Now, let's calculate the force F:

m = 20 kg
g = 9.8 m/s²

Force F = 20 kg * 9.8 m/s² = 196 N

The distance the object is lifted is given as 2.3 m.

Now we have all the information we need to calculate the work done on the object:

Work = Force * Distance * Cos(θ)

However, we need to find the angle θ between the force F and the displacement of the object. Since the object is being lifted vertically, the angle between the force and the displacement is 0 degrees (or 180 degrees). In this case, Cos(θ) = 1.

Work = 196 N * 2.3 m * Cos(0) = 196 N * 2.3 m * 1 = 448.4 J

Therefore, the work done on the object by the force F is 448.4 Joules.