Tricky Problem (at least for me)

[need help]

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...

 

[nrqed]

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 $\mu_k$

 

[kyle8921]

Firstly, to answer the initial question, we can use the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. Plugging in the given values, we get v^2 = (0)^2 + 2(.76)(2) = 3.04. Taking the square root of both sides, we get v = 1.74 m/s as the final velocity.

As for finding F(net), we can use the equation F(net) = ma, where m is the mass and a is the acceleration. In this case, we are given a value for the mass (5kg) and the acceleration (0.76m/s^2). Plugging in these values, we get F(net) = (5)(0.76) = 3.8N.

However, we also need to consider the forces acting on the object, which are the force of gravity, normal force, and force of friction. The net force is the sum of these forces, taking into account their directions. In this case, the force of gravity and normal force are equal in magnitude but opposite in direction, so they cancel each other out. This leaves us with only the force of friction, which is -7.35N. Therefore, the net force is F(net) = -7.35N.

I would advise double checking the given values and equations to ensure accuracy in the calculations. If you are still unsure, it would be best to consult with a teacher or colleague for further clarification.