First, you make a diagram of all the forces acting on the system. This is shown in the figure. We have to determine F1 and F4. Let's do a momentum balance. Momentum is conserved so the summation of all momentum is equal to zero. Momentum is force*distance.
To determine F1: (reference is F4, so F4=0)
∑Momentum = 0 = -F2 - F3 + F1
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.3m)+F1(0.5-0.1m)
F1 = 53.96 N (left knife-edge)To determine F4: (reference is F1, so F1=0)
∑Momentum = 0 = -F2 - F3 + F4
0 = (-4 kg)(9.81 m/s2)(0.25m)-(6kg)(9.81 m/s2)(0.5-0.2m)+F4(0.5-0.1m)
F4 = 68.67 N (right knife-edge)
The question is incomplete. If D=1/T then the answer is D, but if instead of =, there is a proportional sign, then the answer is C.
Answer:
it states that energy can neither be created or destroyed
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:
I'm pretty sure its B and C
Explanation:
B bc the weight is gravitational pull x mass so when the object has same mass the weight is smaller on moon
C bc mass is the same - you can't change it