Question

A roller coaster car of mass 1420 kg slides on a frictionless track starting at a distance 36.83 m above the bottom of a loop 29 m in diameter. The acceleration of gravity is 9.81 m/s 2. If friction is negligible, what is the magnitude of the force of the track on the car when the car is at the top of the loop? Answer in units of N.

Answer 1

Answer:

6859.079 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (39.83-29)+0^2}\\\Rightarrow v=14.57\ m/s$

$N=m\frac{v^2}{r}-mg\\\Rightarrow N=1420\left(\frac{14.57^2}{14.5}-9.81\right)\\\Rightarrow N=6859.079\ N$

Magnitude of the force of the track on the car when the car is at the top of the loop is 6859.079 N

Answer 2

Answer:

Magnitude of force = 6415.207 N

Explanation:

Given that

m= 1420 kg

h= 36.83 m

d=29 m ,r=14.5 m

g=9.81 m/s²          

From energy conservation

Lets take velocity of particle at top position is V

m g h  - m g d =1/2 m V²    

V² = g(h-d)

V²  = 9.81 x (36.83 - 29)

V=8.76 m/s

At top position

F  + m g = mV² /r

$F=\dfrac{mV^2}{r}-mg$

By putting the values

$F=\dfrac{1420\times 8.76^2}{14.5}-1420\times 9.81\ N$

F=−6415.207 N

Magnitude of force = 6415.207 N