Respuesta :
There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: [tex]v=15 m/s[/tex]
- radius of the hill: [tex]r=100 m[/tex]
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car [tex]W=mg[/tex] (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, [tex]m \frac{v^2}{r} [/tex], so we can write:
[tex]mg-N=m \frac{v^2}{r} [/tex] (1)
By rearranging the equation and substituting the numbers, we find N:
[tex]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 [/tex]
(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:
[tex]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 [/tex]
(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:
[tex]mg=m \frac{v^2}{r} [/tex]
from which we find
[tex]v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s [/tex]
- speed of the car at the top of the hill: [tex]v=15 m/s[/tex]
- radius of the hill: [tex]r=100 m[/tex]
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car [tex]W=mg[/tex] (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, [tex]m \frac{v^2}{r} [/tex], so we can write:
[tex]mg-N=m \frac{v^2}{r} [/tex] (1)
By rearranging the equation and substituting the numbers, we find N:
[tex]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 [/tex]
(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:
[tex]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 [/tex]
(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:
[tex]mg=m \frac{v^2}{r} [/tex]
from which we find
[tex]v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s [/tex]
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,
[tex]\rm mg -N=\dfrac{mv^2}{r}[/tex]
[tex]\rm N= mg -\dfrac{mv^2}{r}[/tex] ------ (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,
[tex]\rm N = (975 \times 9.81) - (\dfrac{975\times 15^2}{100})[/tex]
[tex]\rm N = 7371\; N[/tex]
b) Centripetal force is also acting on the driver, therefore
[tex]\rm N'= m_dg -\dfrac{m_dv^2}{r}[/tex] --- (2)
where [tex]\rm m_d[/tex] is the mass of driver.
Now put the values of [tex]\rm m_d,\;g , \; v \;and \;r[/tex] in equation (2) we get,
[tex]\rm N' = (62\times 9.81) - (\dfrac{62\times 15^2}{100})[/tex]
[tex]\rm N' = 469 \; N[/tex]
c) The car speed at which the normal force on the driver equals zero (N =0).This implies that,
[tex]\rm m_d g = \dfrac{m_dv^2}{r}[/tex] --- (3)
Now put the values of [tex]\rm m_d,\;g\;and \;r[/tex] in equation (3) we get,
[tex]\rm 62\times 9.81 = \dfrac{62\times v^2}{100}[/tex]
[tex]\rm v = \sqrt{\dfrac{100\times 62\times 9.81}{62}}[/tex]
v = 31.32 m/sec
For more information, refer the link given below
https://brainly.com/question/13639113?referrer=searchResults
