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Sagot :
Answer:
The man is standing on the plank then he is pulling the rope towards himself. He pushes forward with the legs so that pushes the plank "rope backwards
So that friction f acts in forward direction the plank.
The tension T in the rope is 100 N
[tex](A) \frac{100}{3} N \: towards \: left[/tex]
the fiction force f newtons.
[tex]Man's \: F _{net} ⇒T−F=ma[/tex]
[tex]F _{net} (man)⇒T−F=50a[/tex]
[tex]Plank Fnet ⇒T+f=ma[/tex]
[tex]F _{net} ( plank)⇒T+f=lov \: a[/tex]
[tex]As T=100 N[/tex]
[tex]a = \frac{4}{3} {m/s}^{2} [/tex]
[tex]f = \frac{100}{3} \: newton[/tex]
Answer:
- The mass of man=[tex]50kg[/tex]
- The mass of plank=[tex]100kg[/tex]
Man applies a force of 100 N,therefore,F=100N.
[tex]m=m_o+m_p[/tex]
- Substituting the values, we get.
[tex]m = 50 + 100[/tex]
[tex] \implies{m = 150 kg}[/tex]
Now,
We know that F=[tex]100 N[/tex]
- The formula of the acceleration of system:
[tex] \implies{ \displaystyle{ a = \frac{F}{m} }}[/tex]
[tex] \implies{ \displaystyle{a = \frac{100}{150} }}[/tex]
[tex]\therefore{ \displaystyle{a = \frac{2}{3}m/ {s}^{2} }}[/tex]
Now,let's find the fiction force acting on the man.
[tex]F=m_oa[/tex]
[tex]F=50×\frac{2}{3}[/tex]
[tex]\therefore F=\frac{100}{3} N[/tex]
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