A trailer mechanic pushes a 2500 kg car, home to a speed v, performing a job during the 5000 J. In this process, the car moves 2
5 m. Neglecting friction between the car and roadway:
(A) What is the final velocity v of the car? (B) What horizontal force exerted on the car?
(A) What is the final velocity v of the car? (B) What horizontal force exerted on the car?
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Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,
Therefore, the work done by the worker in lifting the bucket is given as:
Now, plug in the values given and solve for 'W'. This gives,
Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.