A 1500 kg car skids to a halt on a wet road where μk = 0.51. how fast was the car traveling if it leaves 66-m-long skid marks?

We got coefficient of friction uk = 0.51
Mass m = 1500 kg
normal force on the car = weight of the car = 1500 * 9.8 N = 14700 N
and the force of friction on the car, f = uk * normal force = 0.51 * 14700 N = 7497 N
taking Acceleration a = -f/m = -7497/1500 = -4.998 m/s^2
Displacement s = 66 m
Final velocity v = 0
Initial velocity u = ?
v^2 = u^2 + 2as
0 = u^2 - 2 * 4.998 * 66
u = 25.69 m/s

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onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-26T14:41:41+02:00

The speed of the car at the moment it skidded is 25.69 m/s.

The given parameters;

mass of the car, m = 1500 kg
coefficient of friction, μk = 0.51

The frictional force experienced by the car is calculated by applying Newton's second law of motion;

F = ma

The normal force of the car;

Fₙ = mg

Fₙ = 1500 x 9.8

Fₙ = 14700 N

The frictional force of the car is calculated as;

Fk = μk(Fₙ )

Fk = 0.51 x 14700

Fk = 7497 N

The acceleration of the car is calculated as follows;

[tex]a = \frac{F_k}{m} \\\\a = \frac{7497}{1500} \\\\a = 5 \ m/s^2[/tex]

The velocity of the car at the given instant is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\v^2 = 2(5)(66)\\\\v^2 = 660\\\\v =\sqrt{660} \\\\v = 25.69 \ m/s[/tex]

Thus, the speed of the car at the moment it skidded is 25.69 m/s.

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