Physics, 17.07.2019 12:30 Amari1010

A975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road on the car, (b) the normal force exerted by the car on the 62.0-kg driver, and (c) the car speed at which the normal force on the driver equals zero

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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: v=15 m/s

- radius of the hill: r=100 m

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car W=mg

(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

Answer from: pr47723

a) N = 7371 N

b) N' = 469 N

c) v = 31.32 m/sec

Given :

Mass of car = 975 Kg

Mass of driver = 62 Kg

Speed of car = 15 m/sec

Radius of hill = 100 m

Solution :

a) The car is moving in circular motion. Therefore there ia centripetal force acting on the car,

$\rm mg -N=\dfrac{mv^2}{r}$

$\rm N= mg -\dfrac{mv^2}{r}$ ------ (1)

Where, N is normal force exerted by the road acting on the car.

Now put the values of m, g, v and r in equation (1) we get,

$\rm N = (975 \times 9.81) - (\dfrac{975\times 15^2}{100})$

$\rm N = 7371\; N$

b) Centripetal force is also acting on the driver, therefore

$\rm N'= m_dg -\dfrac{m_dv^2}{r}$ --- (2)

where $\rm m_d$ is the mass of driver.

Now put the values of $\rm m_d,\;g , \; v \;and \;r$ in equation (2) we get,

$\rm N' = (62\times 9.81) - (\dfrac{62\times 15^2}{100})$

$\rm N' = 469 \; N$

c) The car speed at which the normal force on the driver equals zero (N =0).This implies that,

$\rm m_d g = \dfrac{m_dv^2}{r}$ --- (3)

Now put the values of $\rm m_d,\;g\;and \;r$ in equation (3) we get,

$\rm 62\times 9.81 = \dfrac{62\times v^2}{100}$

$\rm v = \sqrt{\dfrac{100\times 62\times 9.81}{62}}$

v = 31.32 m/sec

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