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Akimi4 [234]
4 years ago
6

At an accident scene on a level road, investigators measure a car's skid mark to be 98 m long. It was a rainy day and the coeffi

cient of friction was estimated to be 0.40. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.
Physics
1 answer:
shutvik [7]4 years ago
7 0

Answer:

The speed of the car was 28 m/s.

Explanation:

Hi there!

The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:

W = ΔKE = final kinetic energy - initial kinetic energy

In this case, the final kinetic energy is zero, then:

W = - initial kinetic energy

Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:

Wfr = initial kinetic energy

The work done by friction is calculated as follows:

Wfr = Fr · d

Where:

Fr = friction force.

d = distance.

The friction force is calculated as follows:

Fr = N · μ

Where:

N = normal force.

μ = coefficient of friction

Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).

Then the friction force can be written as follows:

Fr = m · g · μ

Where:

m = mass of the car.

g =  acceleration due to gravity (9.8 m/s²)

The work done by friction will be:

W = m · g · μ · d

The equation of kinetic energy is the following:

KE =  1/2 · m · v²

Where

m = mass of the car.

v = speed.

Then:

W = KE

m · g · μ · d = 1/2 · m · v²

2 · g · μ · d = v²

2 · 9.8 m/s² · 0.40 · 98 m = v²

v = 28 m/s

The speed of the car was 28 m/s.

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V{_a}'=-0.95m/s

Explanation:

From the question we are told that:

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Generally the equation for velocity of  A after collision V{_a}' is mathematically given by

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V{_a}'=\frac{4-6.5}{4+M6.5} *4

V{_a}'=\frac{4-6.5}{4+M6.5} *4

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{P_2}-P_1=49.99\ KPa

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d_2=17.5\ cm/s

From continuity equation

A_1v_1=A_2v_2

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v_2=4.21\ m/s

\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2

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\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1

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4 0
3 years ago
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Calculating mass:

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