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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

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

1. and 2. question: in which equation? or i don't understand you, 3. a), 4. c), 5. c), 6. b), 7. d), 8. d) if small number is underlined(behind symbol) , b) if big number is underlined(in front of symbol)9. what is underlined there? 10. c)