help me to solve this..
Answers
m = mo / √ 1-(x^2 /c^2 ),[LHS ]=[M]=[RHS] so 1-(x^2/c^2) is dimensionless .hence (x^2/c^2) is dimensionless .this means that x and c have the same dimensions ,i.e, [x]= [c] = [ LT^ -1]
Answer:
\setlength{\unitlength}{0.9mm}\begin{picture}(5,5)\thicklines\put(0,0){\framebox(10,10){\sf{2\ kg}}}\put(0,5){\vector(-1,0){20}}\put(-26,4){\sf{2a}}\put(10,-0.1){\vector(1,0){10}}\put(21,-1){\sf{f}}\put(5,0){\vector(0,-1){15}}\put(1,-19){\sf{20\ N}}\put(5,10){\vector(0,1){15}}\put(4,26){\sf{R}}\end{picture}
Take \sf{g=10\ m\,s^{-2}.}g=10 ms
−2
.
Since net vertical force is zero, the reaction acting on the block is,
\sf{\longrightarrow R=20\ N}⟶R=20 N
As the truck accelerates forward the block experiences a pseudo force in opposite direction as that of motion of truck.
Thus frictional force is acting on the block opposite to its motion, i.e., along the direction of motion of truck.
The magnitude of frictional force,
\sf{\longrightarrow f=\mu R}⟶f=μR
\sf{\longrightarrow f=0.5\times20}⟶f=0.5×20
\sf{\longrightarrow f=10\ N}⟶f=10 N
Net horizontal force acting on the block (leftwards),
\sf{\longrightarrow 2a_{net}=2a-f}⟶2a
net
=2a−f
\sf{\longrightarrow 2a_{net}=2\times8-10\ }⟶2a
net
=2×8−10
\sf{\longrightarrow 2a_{net}=6\ N}⟶2a
net
=6 N
And net horizontal acceleration,
\sf{\longrightarrow a_{net}=\dfrac{6}{2}}⟶a
net
=
2
6
\sf{\longrightarrow a_{net}=3\ m\,s^{-2}}⟶a
net
=3 ms
−2
The block has to move 6 metres leftwards (along the direction of net acceleration) to fall off the truck.
The time taken to fall off the truck is given by second equation of motion as,
\sf{\longrightarrow s=ut+\dfrac{1}{2}\,a_{net}t^2}⟶s=ut+
2
1
a
net
t
2
Since the block starts from rest,
\sf{\longrightarrow 6=(0)t+\dfrac{1}{2}\,(3)t^2}⟶6=(0)t+
2
1
(3)t
2
\sf{\longrightarrow 6=\dfrac{3}{2}\,t^2}⟶6=
2
3
t
2
\sf{\longrightarrow\underline{\underline{t=2\ s}}}⟶
t=2 s
Hence (3) is the answer.