If an object of 0.550 kg is lifted from the floor to a height of 3.50 m, how much work is done by the lifting force?

What is the kinetic energy?

What is the velocity?

The student's question involves calculating the work done by lifting a mass, as well as the kinetic energy and velocity of the mass at a certain height. Assume the lifting force is equal to the weight of the object and acts in the upward direction, counteracting gravity. The work done (W) by the lifting force can be calculated using the formula W = force × distance, where the force is equal to the weight of the object (mass × acceleration due to gravity), and the distance is the height the object is lifted.
Work Done:
W = m × g × h
Where m is the mass (0.550 kg), g is the acceleration due to gravity (9.8 m/s2), and h is the height (3.50 m).
W = 0.550 kg × 9.8 m/s2 × 3.50 m = 18.815 J (joules)
Kinetic Energy and Velocity:
The question does not specify that the object is moving after being lifted, and if it's simply held at the height of 3.50 m, then the kinetic energy (KE) would be 0, as there is no motion (velocity = 0 m/s). However, if the object were to be dropped from that height without any initial velocity, you could use conservation of energy to predict the velocity just before impact.

Answered by qwstar | 2024-06-18

The work done in lifting a 0.550 kg object to a height of 3.50 m is approximately 18.815 joules. If the object is held at this height without movement, its kinetic energy is 0 joules, and its velocity is 0 m/s. Thus, no kinetic energy has been imparted to the object while it remains stationary at that height.
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Answered by qwstar | 2024-10-02