Advertisement
Not a member of Pastebin yet?
Sign Up,
it unlocks many cool features!
- In 1990 Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth.
- (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant velocity?
- F=m*a
- W=F*d
- Force is constant in this case since the object isn't being affected by anything but his teeth and gravity.
- F=281.5*9.8=2758.7N
- W=F*.171=471.74J
- This is the correct answer
- (b) What total force was exerted on Arfeuille's teeth during the lift?
- ??? It asks for the answer in kiloNewtons. My initial impression was to take the function for Force (in this case it's just the const 2758.7) and to integrate it for the distance .171 meters.
- integrate 2758.7 with respect to x from 0 to .171 = 2758.7x evaluated from 0 to .171
- 2758.7*0.171-0=471.74N = 4.72kN
- This assumes he moves it at 1 meter per second.
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement