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Elan Coil [88]
3 years ago
11

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

uld the car have to accelerate so that a quarter placed on the back wall would remain in place?
Physics
1 answer:
djverab [1.8K]3 years ago
3 0

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

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Answer:

The frictional force is  F_f =  1.75 \  N

Explanation:

From the question we are told that

     The coefficient of kinetic force is  μk = 0.35

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Generally the frictional force that acts on the puck is mathematically represented as

          F_f =  \mu_k  *  F_N

=>       F_f =  0.35  *  5

=>       F_f =  1.75 \  N

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