Question

At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.

Answer 1

force = 11.33
`kg-m/s^(2)`

Step-by-step explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

`P_(1)=mU`

`P_(1)= 17.0 * 4.10 = 69.7 kg- m/s`

from newton second law of motion

`F=(\Delta P)/(\Delta t)`

`F = (P_(1)-P_(2))/(T)`

Kgm/s^2

`F = (69.7-0)/(6.15)= 11.33[tex]kg-m/s^(2)`[/tex]