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

Answer:

force = 11.33 $kg-m/s^{2}$

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

$F=\frac{\Delta P}{\Delta t}$

$F = \frac{P_{1}-P_{2}}{T}$

Kgm/s^2

$F = \frac{69.7-0}{6.15}= 11.33[tex]kg-m/s^{2}$[/tex]