Ask Question
19 October, 20:53

A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi/hr as he passes B, 40 ft along the incline from A. Determine his power output as he approaches B.

+5
Answers (1)
  1. 19 October, 21:19
    0
    88.18 W

    Explanation:

    The weight of the boy is given as 108 lb

    Change to kg = 108*0.453592 = 48.988 kg = 49 kg

    The slope is given as 6%, change it to degrees as

    6/100 = 0.06

    tan⁻ (0.06) = 3.43°

    The boy is travelling at a constant speed up the slope = 7mi/hr

    Change 7 mi/h to m/s

    7*0.44704 = 3.13 m/s

    Formula for power P=F*v where

    P=power output

    F=force

    v=velocity

    Finding force

    F=m*g*sin 3.43°

    F=49*9.81*sin 3.43° = 28.17

    Finding the power out

    P=28.17*3.13 = 88.18 W
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi/hr as he ...” in 📗 Engineering if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers