Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
[tex]F_c = F_g - F_N[/tex]
where;
Fc is the centripetal force
[tex]F_g[/tex] is downward force due to weight of the driver
[tex]F_N[/tex] is upward or normal force on the drive
[tex]F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N[/tex]
Therefore, the normal force the seat exerted on the driver is 125 N.
The normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Given data:
The mass of car is, m' = 2000 kg.
The speed of car is, v = 100 km/h = 100 × 5/18 = 27.77 m/s.
The radius of curvature of path is, r = 100 m.
The mass of driver is, m = 60 kg.
In this case, the normal force on the driver is equal to the difference between weight of the driver and the centripetal force on the driver. Then the expression is given as,
[tex]N'= W - F\\\\N '= mg-\dfrac{mv^{2}}{r}[/tex]
Solving as,
[tex]N' = (60 \times 9.8)-\dfrac{60 \times 27.77^{2}}{100}\\\\N' = 125\;\rm N[/tex]
Thus, we can conclude that the normal force exerted by the seat on the driver if the mass of the driver is 60 kg is of 125 N.
Learn more about the centripetal force here:
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