Force given mass, velocity and fricton

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The discussion focuses on calculating the acceleration and distance traveled by a 2 kg box after an initial push, given a coefficient of friction of 0.20 and an initial speed of 4 m/s. The net force acting on the box is determined to be 3.92 N, leading to an acceleration of 1.96 m/s² when divided by the mass. Participants clarify that the box does not decelerate at 9.8 m/s², as that value represents gravitational acceleration, not friction. To find the distance the box travels before coming to rest, standard kinematic formulas for constant acceleration are recommended. The conversation emphasizes the importance of understanding net forces and applying Newton's second law for accurate calculations.
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Homework Statement


A 2 kg box is given a brief push so that it slides across the floor, but the applied force is only applied for an instant. The coefficient of friction is 0.20 and the initial speed of the box is 4 m/s.
Find acceleration of the box after the push
Find how far the box goes before coming to rest.

Homework Equations


Net force= mass*accel
Friction=u*normal

The Attempt at a Solution


I created an FBD, found the weight is 19.6N(9.8*2), and because it is being pushed horizontally, the Normal force is also 19.6 N. With the normal force I can find the force of friction. F=un, which means F=.2(19.6), and F=3.92 N. This leaves me missing the force I would need to calculate the Net f and find the accleration. I could only get this far, as I do not know how to find a force with only the velocity, mass, and friction.
 
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Hello. Welcome to PF!
Leopard_ said:
Find acceleration of the box after the push
The push was only used to give the object an initial speed of 4 m/s. After that, the pushing force is zero.
 
Oh, I suppose it was easier then I thought! Thank you!
 
So, the box's acceleration of the box is slowing down by 9.81 m/s^2? Because it's initial speed is 4 m/s and there are not other forces acting on it?
 
vr0nvr0n said:
So, the box's acceleration of the box is slowing down by 9.81 m/s^2? Because it's initial speed is 4 m/s and there are not other forces acting on it?
No, the box is not decelerating at 9.8 m/s2. You need to find the net force acting on the box and use Newton's second law of motion to find the acceleration.
 
TSny said:
No, the box is not decelerating at 9.8 m/s2. You need to find the net force acting on the box and use Newton's second law of motion to find the acceleration.
Thank you. I got 1.96 m/s2 by dividing the Net Force by the mass. The Net Force, because this is all on a horizontal plane is the normal force multiplied by the coefficient of friction (3.92 N).

What formula would I need to use to find how far the box goes?

Thank you for your help.
 
vr0nvr0n said:
Thank you. I got 1.96 m/s2 by dividing the Net Force by the mass. The Net Force, because this is all on a horizontal plane is the normal force multiplied by the coefficient of friction (3.92 N).
Yes.

What formula would I need to use to find how far the box goes?
You can use the standard kinematic formulas for constant acceleration.
 

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