Ask Question
10 February, 04:07

A loaded 320 kg toboggan is traveling on smooth horizontal snow at 4.60 m/s when it suddenly comes to a rough region. The region is 8.60 m long and reduces the toboggan's speed by 1.20 m/s. What average friction force did the rough region exert on the toboggan?

+3
Answers (1)
  1. 10 February, 04:41
    0
    367.04 N

    Explanation:

    Frictional force = mass of the toboggan*deceleration caused by the rough region

    Fr = m*ar ... Equation 1

    Where Fr = Average frictional force, m = mass of loaded toboggan, ar = deceleration cause by the rough region.

    We can calculate for ar using the formula below

    v² = u²+2ars ... Equation 2

    make ar the subject of the equation

    ar = (v²-u²) / 2s ... Equation 3

    Where v = final velocity of the toboggan, u = initial velocity of the toboggan, s = distance or length of the rough region

    Given: v = 1.2 m/s, u = 4.6 m/s, s = 8.6 m

    Substitute into equation 3

    ar = (1.2²-4.6²) / (2*8.6)

    ar = (1.44-21.16) / 17.2

    ar = - 19.72/17.2

    ar = - 1.147 m/s²

    Also given: m = 320 kg

    Substitute into equation 1

    Fr = 320 (-1.147)

    Fr = - 367.04 N

    Note: The negative sign tells that the frictional force opposes the motion of the toboggan
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A loaded 320 kg toboggan is traveling on smooth horizontal snow at 4.60 m/s when it suddenly comes to a rough region. The region is 8.60 m ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers