A 1400 kg car is travelling at 25 m/s up a circular hill of radius 210 m. What is the applied force on this car at the top of this hill? (ie. when the car crests the hill)

Asked on by papertissue

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

This is a roller coaster problem where the car moves on the top of the circualr hill, with a radius of R say, The speed of the car is Vv kms.  Taking free  body diagram, the forces on the car when it moves on a curvature of radius R, are the centre petal force, the weight force and the resulting reactional force say T.

Then the net force is zero. So, T- mg  + mv^2/R = 0 , where m is the mass of the car and v its velocity and R the radius of the hill.

T = mg-mv^2/R, Substituting the valus, m =14000kg, v = 25m/s and g = 9.8

= 1400kg(9.8-25^2/210)

= -9553.33 Newton or 9553.33 N towards earth.


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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on


Mass of car = m = 1400 kg

Speed of car = v = 25 m/s

Radius of hill = r = 210 m

we know acceleration due to gravity = g = 9.81 m/s^2


When the car is at the crest of the hill and moving at uniform speed there are two forces acting on it. One is the vertically downward force (f1) due to gravity, and a vertically upward force (f2) due to centrifugal force movement of car along the circular path over the hill.

We calculate the downward gravitational force using the formula:

f1 = m*g = 1400*9.81 = 13734 N

Similarly, we calculate the upward centrifugal force using the formula:

f2 = m*(v^2)/r = 1400*(25^2)/210 = 4166.6667 N

The net force acting on the car is:

= f1 - f2 = 13734 - 4166.6667 = 9567.3333 N downward


Applied force on the car is 9567.3333 N downward.

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