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 long and reduces the toboggan's speed by 1.20 m/s. What average friction force did the rough region exert on the toboggan?

Answer 1

367.04 N

Step-by-step 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