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A 1500kg car is on a road that makes a rather tight right turn. The turn in the road has a radius of 70.0 m. The road is not banked and the coefficient of static friction between the tires and the road is 0.80.

(a) Determine the maximum speed the car can have without sliding off the road. (b) A 2000 kg car takes the same turn. How does its maximum speed compare to that of the 1500 kg car?

2 Answers

3 votes

Final answer:

The maximum speed of a car without sliding off an unbanked road depends on the static friction and the radius of the turn, and is the same for any car regardless of its mass, given the same road conditions. The mass cancels out in the calculation, leading to the conclusion that as long as the coefficient of static friction and road conditions remain the same, both cars can have the same maximum speed.

Step-by-step explanation:

In order to find the maximum speed the car can have without sliding off the road, we need to consider the forces acting on the car. Since there is no banking, the only horizontal force is due to static friction, which acts as the centripetal force required to keep the car moving in a circle. We can use the formula for maximum static friction ℓs max = μs × N, where μs is the coefficient of static friction and N is the normal force (which equals the weight of the car when on flat ground). The centripetal force needed to keep an object moving in a circular path with radius r at speed v is given by Fc = m × v2/r.

For part (a), setting the maximum static friction equal to the required centripetal force gives us μs × m × g = m × v2/r. We can solve for the maximum speed v by rearranging this equation to v2 = μs × g × r. Substituting the given values (μs=0.80, g=9.8 m/s2, r=70.0 m), we can then calculate v.

For part (b), it is important to note that the mass of the car, whether it is 1500 kg or 2000 kg, cancels out in the equation for maximum speed since the mass appears on both sides of the equation. Therefore, the maximum speed for both cars would be the same given the same coefficient of static friction and radius of turn.

User Nouman Zaka
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7.1k points
3 votes

Answer:

v max=4√70 or 33.5m/s,the speed remains the same

Step-by-step explanation:

(a)

m=1500kg

s=70m

coefficient of static friction=0.8

v=?

u=0

R¹=mg

F=ûR

F=0.8×1500×10=8×1500=12000N

F=ma

a=F/m

a=12000/1500

a=8m/s²

finding velocity

we use v²=u²+2as not v=u+at because we don't have time

v²=0²+2×8×70

v²=1120

v=√1120

v max=4√70 or 33.5m/s

(b)

R²=mg=2000×10=20000

Again,

F=ûR

F=0.8×20000=8×2000=16000N

F=ma

16000=2000/a

a=16000/2000=8m/s²

no need to solve the acceleration it is the same then same speed

User Tcxbalage
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8.0k points