W = Fd = 4(2100) = 8400 J
So the answer is A) 8400 J
I was just rewriting my notes on the work lesson I did in class today, so I saw this question at the perfect time!! :)
Hope it helps!! :)
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)
High frequency = D, short wavelength
Answer: 0.62
Explanation:
Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).
Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R
Since W = mg, W = 38.4 ×10
W= 384N =R
Normal reaction = 384N
The horizontal force acting on the body will be the moving force which is 238N
Coefficient of friction = Fm/R
Coefficient of friction = 238/384
Coefficient of friction = 0.62
Therefore, coefficient of kinetic friction between the box and the floor is 0.62
