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Svetlanka [38]
3 years ago
10

John pushes forward on a car with a force of 125n while bob pushes backward on the car with a force of 225n. what is the net for

ce on the car
Physics
2 answers:
Pie3 years ago
5 0

Answer:

100N

Explanation:

because 225-125= 100

Dafna1 [17]3 years ago
5 0

Answer:

200n backwards

Explanation:

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A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient
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Answer:

v_o = 4.54 m/s  

Explanation:

<u>Knowns  </u>

From equation, the work done on an object by a constant force F is given by:  

W = (F cos Ф)S                                   (1)  

Where S is the displacement and Ф is the angle between the force and the displacement.  

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:  

K.E=1/2m*v^2                                       (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:  

W = K.E_f-K.E_o                                 (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020  

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:  

∑F_y=N-mg

     N=mg

Thus, the kinetic friction force is:  

f_k = μ_k*N  

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.  

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:  

W_f=(f_k*cos(180) s

      =-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:  

W= -K.E_o    

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force  

W_f= -K.E_o  

From equation (2), the work done by the friction force in terms of the initial speed is:  

W_f=-1/2m*v^2  

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:  

-μ_k*mg*s = -1/2m*v^2  

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:  

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s  

v_o = 4.54 m/s  

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