Answer
64.8k+ views
Hint: Calculate the net force acting on the vehicle and add it to weight to get the normal force exerted by the track on the vehicle. As the roller coaster is in circular motion the net force on it will contribute to the centripetal acceleration and centripetal force
Complete step by step answer:
The acceleration of the vehicle can be calculated by using the formula
$a = \dfrac{{{v^2}}}{R}$
Where $a$ = acceleration of the vehicle
$v$ = velocity of the vehicle at that point of dip
$R$ = radius of curvature of the dip
Thus,
$a = \dfrac{{{{(18)}^2}}}{{(12)}}$
$a = 27m/{s^2}$
Now the net force acting on the vehicle at the point of dip can be calculated as
${F_{net}} = m \times a$ ,
Where $m$ = mass of the vehicle = $500kg$
${F_{net}} = (500 \times 27)N$
${F_{net}} = 13500N$
${F_{net}} = {F_n} - {F_w}$
At the point of dip, the forces acting on the vehicle are the normal reaction force and the weight of the vehicle exerted by it on the track.
Thus
Weight force of vehicle
${F_w} = mg$
( $g$ is the acceleration due to the gravity )
$ = 500 \times 9.8$
$ = 4900N$
Thus,
Force on the vehicle by the track = ${F_n}$ = $(13500 + 4900)N$
$ = 18400N$
Therefore, the correct answer is option D.
Note: The normal reaction force acts opposite to the direction of exertion of weight force ( towards the ground ). Hence the net force is equal to the difference between the normal reaction force and the weight force. The weight force can also be written as the force of gravity. The normal reaction force is the force exerted by the track on the vehicle whereas the force of gravity is exerted by the vehicle on the track. The net force acts in the upward direction because the movement is circular in nature in which the force acts towards the centre of the circle.
Complete step by step answer:
The acceleration of the vehicle can be calculated by using the formula
$a = \dfrac{{{v^2}}}{R}$
Where $a$ = acceleration of the vehicle
$v$ = velocity of the vehicle at that point of dip
$R$ = radius of curvature of the dip
Thus,
$a = \dfrac{{{{(18)}^2}}}{{(12)}}$
$a = 27m/{s^2}$
Now the net force acting on the vehicle at the point of dip can be calculated as
${F_{net}} = m \times a$ ,
Where $m$ = mass of the vehicle = $500kg$
${F_{net}} = (500 \times 27)N$
${F_{net}} = 13500N$
${F_{net}} = {F_n} - {F_w}$
At the point of dip, the forces acting on the vehicle are the normal reaction force and the weight of the vehicle exerted by it on the track.
Thus
Weight force of vehicle
${F_w} = mg$
( $g$ is the acceleration due to the gravity )
$ = 500 \times 9.8$
$ = 4900N$
Thus,
Force on the vehicle by the track = ${F_n}$ = $(13500 + 4900)N$
$ = 18400N$
Therefore, the correct answer is option D.
Note: The normal reaction force acts opposite to the direction of exertion of weight force ( towards the ground ). Hence the net force is equal to the difference between the normal reaction force and the weight force. The weight force can also be written as the force of gravity. The normal reaction force is the force exerted by the track on the vehicle whereas the force of gravity is exerted by the vehicle on the track. The net force acts in the upward direction because the movement is circular in nature in which the force acts towards the centre of the circle.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)