Question

A 986-kg roller coaster starts from the top of a hill and rolls down. It enters a loop for which the radius at the top is 92.5 meters. The track is designed in such a way that 4260 Newtons is the maximum force allowed in the rail.

Answer 1

Answer:

v = 20 m/s

Explanation:

given,

mass of the roller  = 986 Kg

radius = 92.5 m

maximum allowable force = 4260 N

now, calculating the maximum speed of the roller coaster = ?

now, force due to centripetal acceleration

$F = \dfrac{mv^2}{r}$

total force acting will be equal to

$F =\dfrac{mv^2}{r}$

$4260 =\dfrac{986\times v^2}{92.5}$

$10.659 v^2 = 4260$

       v² = 400

      v = 20 m/s

hence, the maximum speed which roller coaster can attain is equal to v = 20 m/s

Answer 2

Answer:

20 m/s

Explanation:

mass, m = 986 kg

radius, r = 92.5 m

maximum force, F = 4260 N

Let the velocity is v.

the force required is centripetal force.

$F=\frac{mv^{2}}{r}$

$4260=\frac{986\times v^{2}}{92.5}$

v² = 399.65

v = 20 m/s

Thus, the maximum velocity is 20 m/s.