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3 years ago

10

A 1,800 kg car is parked on a road that has an elevation angle of 7°. Suppose the coefficient of static friction of the kinds of

rubber and asphalt involved is 0.65. Which is approximately the force of static friction between the tires and the road? Question options: 11,350 N 9,300 N 18,000 N 2,200 N

2 answers:

pickupchik [31]3 years ago

7 0

Answer:

$F_f = 2200 N$

Explanation:

As we know that car is parked on inclined plane

So here we have

$F_n = mg cos\theta$

$F_n = (1800)(9.81)cos7$

$F_n = 17526.4 N$

now we know that maximum limiting friction between tyres and road is given as

$F_s = \mu_s F_n$

here we have

$F_s = 0.65 \times 17526.4$

$F_s = 11350 N$

Now at static condition of the car net force is balanced on it

so we have

$F_f = mg sin\theta$

$F_f = 1800 \times 9.8 \times sin7$

$F_f = 2200 N$

Masteriza [31]3 years ago

4 0

2,200 N is the ans for the question asked

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m_a_m_a [10]

Hello,

<u>Solution for A:</u>

Force = 3.00N

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Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

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<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

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Time = 3.00 Seconds

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1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

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