The work done by 
along the given path <em>C</em> from <em>A</em> to <em>B</em> is given by the line integral,
I assume the path itself is a line segment, which can be parameterized by
with 0 ≤ <em>t</em> ≤ 1. Then the work performed by <em>F</em> along <em>C</em> is
![\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cleft%286x%28t%29%5E3%5C%2C%5Cvec%5Cimath-4y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%29%5Ccdot%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Bx%28t%29%5C%2C%5Cvec%5Cimath%20%2B%20y%28t%29%5C%2C%5Cvec%5Cjmath%5Cright%5D%5C%2C%5Cmathrm%20dt%20%5C%5C%5C%5C%20%3D%20%5Cint_0%5E1%20%28288%283t-1%29%5E3-8%282t%2B5%29%29%20%5C%2C%5Cmathrm%20dt%20%3D%20%5Cboxed%7B312%7D)
I’m not really sure I’m sorry
<h2>The man have to apply force of 160 N</h2>
Explanation:
The work done to lift the bag of weight mg through height 2.5 m is 400 J
The work done can be found by relation W = mg x h
Thus mg = 
= 
= 160 N
Therefore the man have to apply the force of 160 N
"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."
These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way. So the way WE express and use the 2nd law of motion is
"<em>F = m·A.</em> The net force on an object is equal to the product of the object's mass and its acceleration."
The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.
