an object is thrown vertically upward from the earth;s surface with certain initial speed. it rises to a maximum height and then falls back to the starting point. what is the wrok done by the by the gravitational force in the process?exaplain

 

 

ANSWER is

the total work done by the gravitational force is zero as work is done against the gravitational force by the object during the upward journey. and the same magnitude of work is done on the object in the downard journey

I noticed this thread in a forum.

However, my answer does not reconcile this one.

I think since v^2 is a vector and it changed, hence KE is changed since magnitude is unchanged but direction is changed,

But my teacher taught me KE is not a vector, so what's the correct answer?

velocity is a vector, speed isn't

The scalar product velocity.velocity = speed² because Cos(0)=1 making KE not a vector.

Work done for a constant force is

[math] W= \vec{F} \cdot \vec{s} [/math]

where [math]\vec{F}[/math] is the force and [math]\vec{s}[/math] is the displacement.

 

On the way up, [math]\vec{F} = -mg \vec{j}[/math] and [math]\vec{s} = h \vec{j}[/math] where 'h' is the height it reaches and [math]\vec{j}[/math] is the unit vector in the 'up' direction, so

[math]W_{\rm up} = -mgh[/math]

 

On the way down, [math]\vec{F} = -mg \vec{j}[/math] as before but [math]\vec{s} = -h \vec{j}[/math], so

[math]W_{\rm down} = mgh[/math]

 

The total work done is [math]W_{\rm up}+W_{\rm down} = 0.[/math]

 

Alternatively, the work done (for a varying force) is:

[math] W = \int \vec{F} \cdot d\vec{s} = \int_0^h (-mg) dy + \int_h^0 (-mg) dy = mgh-mgh=0[/math].