Tricky Problem (at least for me)

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To find the final velocity with an initial velocity of 0 m/s, an acceleration of 0.76 m/s², and a distance of 2 meters, the equation v_f² = v_i² + 2a(x_f - x_i) can be used, resulting in a final velocity of approximately 2.47 m/s. For calculating the net force (F_net), the force of friction is determined by multiplying the normal force (-49 N) by the coefficient of friction (0.15), yielding -7.35 N. F_net can then be calculated by summing the forces acting on the object, including gravity and friction. The total force acting on the object, considering its mass and acceleration, is essential for determining F_net. Understanding these relationships is crucial for solving the problem accurately.
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If I have a initial velocity of 0m/s and an acceleration of .76m/s2 and a distance of 2meters what would my final velocity be?


Also I need to find F(net) I know a mu of .15 a time of 5seconds a mass of 5kg and force of gravity equals 49N an normal force equals -49N and force of friction equals -7.35N. I think I'm supposed to use the velocity found in the question above, but I'm not sure and I don't know how. How do I find F(net) with the information I have and what is the answer??

Any help would be greatly appriciated...
 
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need help said:
If I have a initial velocity of 0m/s and an acceleration of .76m/s2 and a distance of 2meters what would my final velocity be?
v_{f,x}^2 = v_{i,x}^2 + 2 a_x (x_f-x_i) where x_f-x_i is essentially the distance traveled. have you seen this equation before?
Also I need to find F(net) I know a mu of .15 a time of 5seconds a mass of 5kg and force of gravity equals 49N an normal force equals -49N and force of friction equals -7.35N. I think I'm supposed to use the velocity found in the question above, but I'm not sure and I don't know how. How do I find F(net) with the information I have and what is the answer??

Any help would be greatly appriciated...
It

The key point is that the magnitude of the friction force is equal to the magnitude of the normal force times the coefficient of kinetic friction \mu_k
 
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