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poizon [28]
3 years ago
14

What is the maximum speed at which a car can safely travel around a circular track of radius 75.0 m if the coefficient of fricti

on between the tire and road is 0.200?
A) 3.87 m/s
B) 12.1 m/s
C) 15.0 m/s
D) 147 m/s
Physics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer:

B) 12.1 m/s

Explanation:

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the radial direction:

∑F = ma

F = m v² / r

Nμ = m v² / r

Substituting and solving for v:

mgμ = m v² / r

gμ = v² / r

v = √(gμr)

Given that μ = 0.200 and r = 75.0 m:

v = √(9.81 m/s² × 0.200 × 75.0 m)

v = 12.1 m/s

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A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

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