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3 years ago

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A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F

ind the minimum stopping distance so the computer does not slip and fall onto the floor if the coefficient of static friction between the seat and the laptop is 0.40 and the coefficient of kinetic friction is 0.20.

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer: $32.65\ m$

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is $\mu_s=0.4$

The coefficient of kinetic friction is $\mu_k=0.2$

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction $\mu_kg=0.2\times 9.8\ m/s^2$

Using the equation of motion $v^2-u^2=2as\\$

insert the values

$0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m$

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