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# A car moves at a constant speed on a straight but hilly road. One section has a crest and dip of the same 250 m radius. As the car passes over the crest the normal force on the car is one half the 16kN weight of the car. What will be the force on the car as its passes through bottom of the dip:A. 12kNB. 24kNC. 36kND. 48kN

Last updated date: 15th Sep 2024
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Hint:-Whenever anybody passes through the curve centripetal force acts on the body and the direction of the force is at the body and is always acting away towards the centre of the curve. The centripetal force is necessary to act on the body as the curve arises as to balance the forces on the body.
Formula used: The formula of the centrifugal force is given by,
${F_c} = \dfrac{{m{v^2}}}{r}$
Where ${F_c}$is the centripetal force m is the mass v is the velocity of the body and r is the radius of the curve.
The formula of weight is given by,
$w = mg$
Where w is the weight m is the mass and g is the acceleration due to gravity.

Complete step-by-step solution
It is given in the problem that a car moves at a constant speed on a straight but hilly road one section of the hilly road is crest and dip of the same radius equal to 250 m as the car passes over the crest the normal force on the car is one half the 16kN weight of the car and we need to find out the normal force when the car goes from the dip of same radius.
Let us consider the case when the car goes from the crest.

Here we can observe that the weight of the body and the centripetal force both are acting towards the centre of the curve but the normal force is acting in the opposite direction.
$\Rightarrow mg - {N_1} = \dfrac{{m{v^2}}}{r}$
Also it is given in the problem that normal force is half of the weight of the car.
$\Rightarrow mg - \dfrac{{mg}}{2} = \dfrac{{m{v^2}}}{r}$
The value of normal force is half of weight and is equal to 16kN.
$\Rightarrow \dfrac{{mg}}{2} = \dfrac{{m{v^2}}}{r} = 16kN$………eq. (1)
Now let us consider the case when the body is moving through the dip.

Here we can observe that the centrifugal force is acting in the same direction as the normal force and weight is in the opposite direction.
$\Rightarrow {N_2} - mg = \dfrac{{m{v^2}}}{r}$
The value of $\dfrac{{m{v^2}}}{r}$ is equal to $\dfrac{{mg}}{2}$ from the equation (1)
$\Rightarrow {N_2} - mg = \dfrac{{m{v^2}}}{r}$
Replace $\dfrac{{m{v^2}}}{r} = \dfrac{{mg}}{2}$ in the above relation.
$\Rightarrow {N_2} - mg = \dfrac{{mg}}{2}$
$\Rightarrow {N_2} = \dfrac{{mg}}{2} + mg$
$\Rightarrow {N_2} = \dfrac{{3mg}}{2}$
Solving we get,
$\Rightarrow {N_2} = \dfrac{{3 \cdot \left( {16} \right)}}{2}$
$\Rightarrow {N_2} = 24kN$.
The normal force acting on the car as it passes from the dip is equal to${N_2} = 24kN$. The correct answer for this problem is option B.

Note:- The centripetal force is the same force that keeps the vehicles to stay on the roads while taking the turn on the circular track. The centripetal force depends upon the mass, velocity and radius of the cure on which body is moving.