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4 years ago

5

A roller-coaster car has a mass of 1040 kg when fully loaded with passengers. As the car passes over the top of a circular hill

of radius 24 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v?

1 answer:

rusak2 [61]4 years ago

8 0

Answer:

a.6373.5 N

b.-3837.6 N

Explanation:

Mass of roller coaster=m=1040 kg

Radius=r=24 ,

a. $v=9.4m/s$

Normal force= $F_N$

According to question

$mg-F_N=\frac{mv^2}{r}$

Where $g=9.81 m/s^2$

Substitute the values

$1040\times 9.81-F_N=\frac{1040\times (9.4)^2}{24}$

$10202.4-F_N=3828.9$

$F_N=10202.4-3828.9=6373.5 N$

b. $v=18m/s$

$g=9.81 m/s^2$

$1040\times 9.81-F_N=\frac{1040\times (18)^2}{24}$

$10202.4-F_N=14040$

$F_N=10202.4-14040=-3837.6 N$

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