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erma4kov [3.2K]
3 years ago
9

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.
Physics
1 answer:
jasenka [17]3 years ago
8 0

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]

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zimovet [89]

Answer:

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          .v = RA 2 e DV / m

          .v = RA (2 1.6 10-19 51400 / 9.1 10-31)

           .v = RA (1.8075 10 16)

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Answer:

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