Physics High School

A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radius of 40.0 m. The coefficient of static friction between the car tires and the road is 0.500. The car travels successfully around the turn at a constant speed of 10.0 m/s. Calculate the magnitude of the car's acceleration as it goes around the turn. _______ m/s^2

Answers

Answer 1

Answer:

2.5 m/s²

Explanation:

The given parameters are;

The mass of the car, m = 2,000 kg

The radius of the car, r = 40.0 m

The coefficient of friction between the car tires and the road, μ = 0.500

The constant speed with which the car moves, v = 10.0 m/s

The normal reaction of the road on the car, N = The weight of the car;

∴ N = m × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

N ≈ 2,000 kg × 9.81 m/s² = 19,620 N

The frictional force, [tex]F_f[/tex] = μ × N

The centripetal force, [tex]F_c[/tex] = m·v²/r

The car moves without slipping when [tex]F_f[/tex] = [tex]F_c[/tex]

Therefore, [tex]F_f[/tex] = 0.500 × 19,620 N = 2,000 kg × [tex]v_{max}[/tex]²/40.0 m

∴ [tex]v_{max}[/tex] = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s

Therefore, the velocity with which the car moves, v < [tex]v_{max}[/tex]

The cars centripetal acceleration, [tex]a_c[/tex] = v²/r

∴ [tex]a_c[/tex] = (10.0 m/s)²/40.0 m = 2.5 m/s²

The cars centripetal acceleration as it goes round the turn, [tex]a_c[/tex] = 2.5 m/s².

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pf

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