Block pulled up the incline problem

[rudransh verma]

Homework Statement

A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases
to act. Take g = 10 m/s

Relevant Equations

$v=u+at$
$W=F.d$
$v^2=u^2+2as$

a) $$v=u+at$$
$$v=0+(F/m)t$$
$$v=-10$$
Now $$v^2=0+2as$$
$$s=-5$$
$$W=F.d=(-20)(-5)=100J>40$$

b) $$W=-mgh$$
$$W=-20(3)$$
$$W=-60$$

 

[Tea_Aficionado]

Homework Statement:: A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases
to act. Take g = 10 m/s
Relevant Equations:: $v=u+at$
$W=F.d$
$v^2=u^2+2as$

a) $$v=u+at$$
$$v=0+(F/m)t$$
$$v=-10$$
Now $$v^2=0+2as$$
$$s=-5$$
$$W=F.d=(-20)(-5)=100J>40$$

b) $$W=-mgh$$
$$W=-20(3)$$
$$W=-60$$

In my humble opinion, I'd suggest checking your setup for acceleration. What other forces are there acting upon the "x" (parallel to the plane) direction?

 

[Lnewqban]

The F that you use in your equation should be the net force, or addition of all the forces that are parallel to the incline.
Forces acting in opposite direction to the given 20 N do not allow the block to reach the values of velocity and distance that you have calculated.

 

[rudransh verma]

The F that you use in your equation should be the net force, or addition of all the forces that are parallel to the incline.

Yes! Parallel force!
That is only the applied 20N force
Maybe F_N is slowing down the block but that is not parallel to the incline plane.

 

[Lnewqban]

Yes! Parallel force!
That is only the applied 20N force
Maybe F_N is slowing down the block but that is not parallel to the incline plane.

What is the reason for the incline to be present in this problem?
Why it is harder to ride a bicycle uphill?

 

[rudransh verma]

Why it is harder to ride a bicycle uphill?

Because of the Weight of the bicycle

 

[Lnewqban]

Because of the Weight of the bicycle

Exactly!
Where is the weight of the block being considered in your calculations?

 

[rudransh verma]

Exactly!
Where is the weight of the block being considered in your calculations?

But how can we find net force if weight is downward and applied force is parallel to the incline?

 

[Lnewqban]

The weight of a drop of water is also downward, and still, it makes a river flow.

 

[phinds]

But how can we find net force if weight is downward and applied force is parallel to the incline?

Would it matter if the incline were 10 degree instead of 37 degrees? How about if it were 75 degrees?

 

[Tea_Aficionado]

But how can we find net force if weight is downward and applied force is parallel to the incline?

I think what Lnewqban is getting at is that we should split the force due to gravity into "x" (direction parallel to incline) and "y" (direction perpendicular to incline). I noticed in some of your earlier threads that you err in the pattern of thinking that mg is always perpendicular to motion (which is a reasonable error, because we're usually initially taught in horizontal planes).

We want to split mg into these components because I think we want to always consider forces in the direction of motion. After all, F = ma, and it's much easier to solve for acceleration if F is in the same direction!

EDIT: Lnewqban beat me to the diagram :p

 

[haruspex]

Homework Statement:: A block of mass 2.0 kg kept at rest on an inclined plane of inclination 37° is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J.

I hope the original specified frictionless.

Take g = 10 m/s

$10m/s^2$.

 

[Lnewqban]

Please, see:
https://www.physicsclassroom.com/class/vectors/Lesson-3/Inclined-Planes

 

[rudransh verma]

Where is the weight of the block being considered in your calculations?

a) Actually I took the x-axis parallel to horizontal when resolving components of forces and when calculating distance I took x parallel to incline which is not consistent. Right?
$v=(12-20)/2$
$v=-4$
$16=2(-4)s$
$s=-2$
$W=-20(2)=-40J$(edit)

b) $W=-mgh=-20(1.2)=-24J$

c) $KE=(1/2)mv^2$
$KE=16J$
At the moment the force ceases to exist the KE will be 16J

 

[haruspex]

-20(-2)=-40J

Try again.

Thinking a bit more about my post #12 and the wordings "that the work done by the applied force does not exceed 40 J" and "if the work done by the applied force is 40 J", I suspect that 1) it is supposed to say "that the work done by the applied force is at least 40 J" and 2) you are supposed to allow for the possibility of friction.

 

[rudransh verma]

I suspect that 1) it is supposed to say "that the work done by the applied force is at least 40 J"

Yeah

you are supposed to allow for the possibility of friction.

The question doesn’t talk about friction. So the work done by non conservative force is zero. If we use the eqn $W_{nc}+W_{ext}=\Delta E$ , $W_{nc}=0$
I have a bigger concern. I am making repeated mistakes in problems of incline plane because of the non consistent use of xy orientation. To be clear that I got it this time the x orientation parallel to incline is taken when we also deal with suvat eqns up or down the incline. We resolve our forces in that orientation only. But if say we have a body which is colliding with the incline plane like my recent problem we need to take x-axis parallel to horizontal.

 

[haruspex]

like my recent problem we need to take x-axis parallel to horizontal.

You can take the axes however you like, though it could be quite confusing if they're not orthogonal. For some problems it’s easier using horizontal and vertical, for others, parallel and normal to a surface.

The question doesn’t talk about friction.

Yes, and that bothers me. If you are supposed to ignore friction then it ought to say so, and there would be no reason to specify "does not exceed".

 

[nasu]

The work is 40 J in absence of friction. You are supposed to show that in presence of friction the work is less than 40 J. You don't need to know how much is the friction to solve (a). So, the work is at most 40 J. (Not at least).

 

[haruspex]

The work is 40 J in absence of friction. You are supposed to show that in presence of friction the work is less than 40 J. You don't need to know how much is the friction to solve (a). So, the work is at most 40 J. (Not at least).

Thanks for correcting me. Hadn’t thought through it properly.

 

[rudransh verma]

Thanks for correcting me. Hadn’t thought through it properly.

The question can simply ask “how much work is done by the applied force?”. Because if everything is constant like a,u,t then W will be constant too. There is only one value of W=40J.

 

[haruspex]

The question can simply ask “how much work is done by the applied force?”. Because if everything is constant like a,u,t then W will be constant too. There is only one value of W=40J.

You are not taking it in.
@nasu and I agree that the question intends you to allow for the possibility of friction, and to show that even if there is friction the work done by the force is at most 40J.
Do you want to get this question right or not?

 

[rudransh verma]

and to show that even if there is friction the work done by the force is at most 40J.

You mean $W=20\frac{\mu F_N-8}4<=40J$.

 

[haruspex]

You mean $W=20\frac{\mu F_N-8}4<=40J$.

Not quite.
First, you have a sign error. $W=20\frac{\mu F_N-8}4$ would be >=-40J.
Secondly, it would be better to use $\mu_k$, indicating kinetic friction.
Thirdly, you should consider what happens if $\mu_k F_N>8 N$

How do you know that? They did not talk about friction anywhere in the question.

Precisely. It does not say there is no friction. At the level you are at, you should consider friction unless it tells you not to. (This is different from aerodynamic drag, where usually you will not be expected to consider it.)
More tellingly, the question only asks you to show the work is not more than 40J. If you were meant to ignore friction, surely it would ask you to show it is exactly 40J.