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)
Answer:
The ball will fall on the X .
Explanation:
At height, when the aeroplane is in great speed , everything attached with it acquires the same speed . So ball will also have the same speed as the aeroplane have. When ball starts falling off , it gets detached from plane but , at the same time it continues to travel with its earlier speed , because of inertia of motion. So it remains stationary with respect to plane in horizontal direction . It has velocity with respect to plane only in vertical direction. Hence it will fall on the X. It is due to first law of motion.
Answer:
182.28 W
Explanation:
Here ,
m = 7.30 Kg
distance , d= 28.0 m
time , t = 11.0 s
average power supplied = change in potential energy/time
average power supplied = m×g×d/time
average power supplied = 7.30×9.81×28/11
average power supplied = 182.28 W
the average power supplied is 182.28 W
