Calculate the force F necessary to keep the velocity constant.

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To calculate the force F necessary to keep a 10-kilogram block moving at a constant velocity of 6.0 m/s on a rough surface with a coefficient of friction of 0.2, the force of friction must be determined first, which is 20 Newtons. Therefore, the applied force F must equal the friction force to maintain constant velocity, resulting in F being 20 Newtons. For part B, when the block accelerates and its kinetic energy increases by 60 joules, the work-energy theorem indicates that the net work done equals the change in kinetic energy, requiring a different force F' to account for this increase. The discussion emphasizes the distinction between force and energy, clarifying that they cannot be equated directly. Understanding these principles is essential for solving the problems accurately.
AraProdieur
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Ok, so here's the problem.
A 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F as shown above. At time t=0, the velocity v of the block is 6.0 meters per second in the same direction as the force. The coefficient of sliding friction is 0.2. Assume g= 10 meters per second squared.
A) Calculate the force F necessary to keep the velocity constant.

The force is now changed to a larger constant value F'. The block accelerates so that its kinetic energy increases by 60 joules while it slides a distance of 4.0 meters.

B) Calculate the force F'.

C) Calculate the acceleration of the block.


Ok, so far I have that the work done by kinetic energy is 1/2mv^2 by which would be simplified to 180 Joules. Since there is friction involved, I would account for the Force of friction which is the coefficient of friction multiplied by the normal force in which the total would be 2.
Now would the force F be 180 Joules to keep it constant?

and for B since the kinetic energy is increased by 60 Joules, then the total kinetic energy put on the object would be 240 Joules. So would that be force F'?

and for the acceleration of the block, would I use V^2= Vo^2 + 2a(deltax)?
 
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AraProdieur said:
Ok, so here's the problem.
A 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F as shown above. At time t=0, the velocity v of the block is 6.0 meters per second in the same direction as the force. The coefficient of sliding friction is 0.2. Assume g= 10 meters per second squared.
A) Calculate the force F necessary to keep the velocity constant.

The force is now changed to a larger constant value F'. The block accelerates so that its kinetic energy increases by 60 joules while it slides a distance of 4.0 meters.

B) Calculate the force F'.

C) Calculate the acceleration of the block.


Ok, so far I have that the work done by kinetic energy is 1/2mv^2 by which would be simplified to 180 Joules. Since there is friction involved, I would account for the Force of friction which is the coefficient of friction multiplied by the normal force in which the total would be 2.
the normal force is mg which is 100 Newtons, so the friction force is 20 Newtons.
Now would the force F be 180 Joules to keep it constant?
you are confusing force with energy. What does Newton 1 tell you about objects moving at constant velocity? What then must be the applied force F?
and for B since the kinetic energy is increased by 60 Joules, then the total kinetic energy put on the object would be 240 Joules. So would that be force F'?
you can't equate force and energy, they are 2 different animals. Are you familiar with the work-energy theorem, which states that the work done by all forces acting on the block must be equal to the change in Kinetic energy of the block?
 
So then how would I go about solving these problems?
 
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