Question

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A force of 2,500 pounds drags a truck 30 feet in 1/2 minute.

How much work is accomplished?

How much power was required?

Work = 75,000 ft-lbs, Power = 32, 5000 ft-lbs/sec
Work = 7,500 ft-lbs, Power = 250 ft-lbs/sec
Work= 75,000 ft-lbs, Power = 2,500 ft-lbs/sec

Answer 1

Work= 75,000 ft-lbs, Power = 2,500 ft-lbs/sec

Step-by-step explanation:

Answer 2

In order to calculate the work done on a body, we find the product of the force applied on the body and the displacement produced in the direction of the force. Mathematically:
Work = Force * displacement
Work = 2,500 * 30
Work = 75,000 ft-lbs

Power is the rate of work done, this is:
Power = Work / time
Power = 75,000 / 30
Power = 2,500 ft-lbs/sec

Work= 75,000 ft-lbs, Power = 2,500 ft-lbs/sec