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 passes B, 40 ft along the incline from A. Determine his power output as he approaches B.

Answer 1

Answer:

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