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creativ13 [48]
3 years ago
6

A pillow with mass of 0.3 kg sits on a bed with a coefficient of static friction of 0.6. What is the maximum force of static fri

ction?
Physics
2 answers:
lys-0071 [83]3 years ago
5 0
The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body. 
 
                        P = W = mass x acceleration due to gravity
    
                    P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N

Solving for the static friction force (F), 
                                              F = P x c 
 
                                      F = (2.94 N) x 0.6 = 1.794 N

Therefore, the maximum force of static friction is 1.794 N. 



bazaltina [42]3 years ago
5 0

Answer:

1.8 N

Explanation:

A P E X verified

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(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

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Solving for F,

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(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

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