Work done in moving a body up an incline

[rudransh verma]

Homework Statement

A block of 100N weight is slowly slid up on a smooth incline of inclination 37 degrees by a person. Calculate the work done by the person in moving the block through a distance of 2m, if the driving force is a) parallel to the incline and b) in the horizontal direction.

Relevant Equations

$W=F.d$

$W=mgh=100(\sin 37)2=-120J$ Right answer!
But the question is asking work done by the person. So again I wrote two eqns
$F_N\sin 53+F_D\sin 37-100=10.2a_y$
$F_N\cos 53-F_D\cos 37=-10.2a_x$
I just need $a_x$ and $a_y$ to solve.

 

[BvU]

Did you check your dimensions ?

"slowly" means $a=0$.

$\$

 

[rudransh verma]

Did you check your dimensions ?

"slowly" means $a=0$.

$\$

If net force is zero then work done by the person W=238.09J
Where have I done wrong?

 

[vcsharp2003]

If net force is zero then work done by the person W=238.09J
Where have I done wrong?

Isn't work done by external force(s) excluding gravity equal to the change in mechanical energy of the system i.e. block in this case? Work must manifest as an increase or decrease of an equal amount of energy.

 

[BvU]

If net force is zero then work done by the person W=238.09J
Where have I done wrong?

No idea. You don't tell us what you are doing, you only give us a numerical result.

[edit] did you notice there is a part a) and a part b) in this exercise ?

 

[rudransh verma]

Isn't work done by external force(s) excluding gravity equal to the change in mechanical energy of the system i.e. block in this case?

Yeah! But we can simply use $W=F.d=F_D.d=-119.04*-2=238.09J$

 

[BvU]

How is work defined ? Which way is the force ?

 

[rudransh verma]

How is work defined ? Which way is the force ?

a) says parallel to incline. So $F_D$ is anti parallel to positive x-axis as well as displacement. So both will be -ve.
$W=\vec F.\vec d$

 

[vcsharp2003]

Yeah! But we can simply use $W=F.d=F_D.d=-119.04*-2=238.09J$

I would think that the shortest and easiest way to solve this problem would be to find the change in total mechanical energy.

You could alternately solve it by determining the applied force by person, but it would involve more involved calculations.

 

[vcsharp2003]

Homework Statement:: A block of 100N weight is slowly slid up on a smooth incline of inclination 37 degrees by a person. Calculate the work done by the person in moving the block through a distance of 2m, if the driving force is a) parallel to the incline and b) in the horizontal direction.
Relevant Equations:: $W=F.d$

W=mgh=100(\sin 37)2=-120J Right answer!
But the question is asking work done by the person. So again I wrote two eqns
$F_N\sin 53+F_D\sin 37-100=10.2a_y$
$F_N\cos 53-F_D\cos 37=-10.2a_x$
I just need $a_x$ and $a_y$ to solve.

For incline plane problems, wouldn't it be more convenient to take x-axis as parallel to the incline and y-axis as perpendicular to the incline? You have taken the horizontal axis as x-axis and the vertical axis as y axis.

 

[rudransh verma]

I would think that the shortest and easiest way to solve this problem would be to find the change in total mechanical energy.

Then $W_{ext}=\Delta E=U_f+K_f-U_i-K_i=U_f-U_i=mgh=100(1.2)=120J$
Same for b)

I have two questions
1) Is normal force not a non conservative force?
2) why W=F.d doesn’t work here?

 

[BvU]

1) no
2) it works perfectly well

 

[rudransh verma]

2) it works perfectly well

But $W=F_D.d= -119.04*-2=238.09 J$.
By the way I have taken standard orientation of xy axis along the horizontal and vertical. But when I am using the formula for work I am taking incline as my x axis.

 

[rudransh verma]

it works perfectly well

Got it! Some blunder in calculation W=120J