Question

A block of mass m is placed on a floor of lift moving with velocity 4t*2.find the time at ehich normal force on the block is three times of its weight

Answer 1

The velocity of the block is,

$\longrightarrow\sf{v=4t^2}$

So the acceleration of the block is,

$\longrightarrow\sf{a=\dfrac{dv}{dt}}$

$\longrightarrow\sf{a=\dfrac{d}{dt}\left(4t^2\right)}$

$\longrightarrow\sf{a=8t}$

For the normal force on the block be 3 times its weight, the lift should move upward, because the block experiences gain in weight when lift moves upward.

The free body diagram of the block is given below.

$\setlength{\unitlength}{0.15cm}\begin{picture}(5,5)\thicklines\put(0,0.01){\framebox(6,6){\sf{m}}}\put(3,0){\vector(0,-1){10}}\put(1.7,-12){\sf{mg}}\put(3,6){\vector(0,1){10}}\put(2.4,17){\sf{R}}\put(10,-2){\vector(0,1){10}}\put(11.5,2.5){ \sf{a=8t} }\end{picture}$

The net force acting on the block is,

$\longrightarrow\sf{ma=R-mg}$

$\longrightarrow\sf{8mt=R-mg}$

So the normal force on the block is,

$\longrightarrow\sf{R=mg+8mt}$

$\longrightarrow\sf{R=m(g+8t)}$

For the normal force be 3 times its weight,

$\longrightarrow\sf{R=3mg}$

$\longrightarrow\sf{m(g+8t)=3mg}$

$\longrightarrow\sf{g+8t=3g}$

$\longrightarrow\sf{8t=2g}$

$\longrightarrow\underline{\underline{\sf{t=\dfrac{g}{4}}}}$

Taking $\sf{g=9.8\ m\,s^{-2},}$

$\longrightarrow\underline{\underline{\sf{t=2.45\ s}}}$

Therefore, normal force on the block is 3 times its weight at 2.45 seconds.