Starting with one of Newton\'s laws and your free body diagram, derive an expres

Question

Starting with one of Newton's laws and your free body diagram, derive an expression for the normal force exerted on the block by floor. Starting with one of Newton's laws and your free body diagram, derive an expression for the coefficient of kinetic friction Create a qualitative graph of the speed v of the block as a function of time t. Create a qualitative graph of the position x of the block as a function of time t. Derive and expression for the magnitude of the force F1 when the block loses contact with the surface. Call this quantity Derice an expression for the magnitude of the greatest acceleration Image the mass of the block doubles. What does this do to your answers for Remember to answer with:

Explanation / Answer

1.Always remember,normal reaction is no constant force.It adjusts itself accordingly to make the body come in euilibrium.Hence,N=mg-F1sin theta.When theta is 0,the F1 does not affect the N since the two are perpendicular.So

N=mg.

2.Frictional force is the force the floor exerts for opposing the movement of the block.Hence uk N= F1cos theta.Hence

uk=Fcos theta/mg - F1sin theta. uk doesnt have a unit coz its force/force.

3.We know that a constant force and hence a constant acceleration acts in the forward direction on the flock=Fcos theta/m.Now,we know that acceleration is dv/dt.So the velocity time graph can be found out by integrating the above with t.So we get, v= Fcos theta t/m.So the graph is a linear one.

4.For the position time graph ,since the motion is in a straight line,you can directly integrate the vel-time graph wrt time.So,we get x= Fcos theta t^2/2m.Hence it is a parabola in the first quadrant extending upwards.

5.When the block loses contact with the surface,the force which is adjustable and which depends on the contact with the surface will vanish i.e, the normal reaction.Hence, F1max sin theta=mg.So F1max=mg/sin theta.

6.Now we know F1max.So,the acceleration in the forward direction is Fimax cos theta/m=gcot theta.

7.If m doubles,F1 max doubles as you can see from the equation and a is unaffected as tgere is no m term in its equation.Yes it does,if you have a heavier body,you need more force to lift it but the acceleration it has is not affected.Think it this way,you can acclerate a truck and a car both to the same amount.Its just that the force required for the truck is more!.Couldnt provide graphs but i hope the concept would be clear.All the best!

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