Question

A box of mass 70 kg is pulled by a horizontal force of 500 N on the surface of the floor.
When the box moves, the co-efficient of friction between the floor and the box is 0.50.
Calculate the acceleration of the box.

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}}$

• Given :

➡ mass of block = 70kg

➡ Fext = 500N

➡ co-efficient of friction = 0.5

• To Find :

➡ Acceleration produced in block by net force

• Formula :

➡ Frictional force is given by

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \blue{ \bold{ \sf{f = \mu \times mg}}}}$

➡ Net force Fnet is given by

$\: \: \: \: \: \boxed{ \blue{ \bold{ \sf{Fnet = Fext - f = Fext - \mu \: mg}}}}$

➡ As per Newton's second law of motion

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \blue{ \bold{Fnet = m \times a}}}}$

• Calculation :

➡ Fnet = 500 - 0.5(700) = 500-350 = 150N

➡ Fnet = ma

➡ a = 150/70 = 2.14 m/s^2

$\: \: \: \: \: \: \: \: \: \: \star \: \boxed{ \boxed {\sf{ \green{ \bold{acc. = 2.14 \frac{m}{ {s}^{2} }}}}}} \: \star$

Answer 2

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put (9,0){\framebox(10,10){\sf{70 kg}}}\put(14,0){\vector(0,-1){10}}\put(9.6,-13.5){\sf{700\ N}}\put(14,10){\vector(0,1){10}}\put(13,21.5){\sf{R}}\put(-20,0){\line(1,0){68}}\multiput(19,5)(5,10){2}{\vector(1,0){10}}\put(30,4){\sf{500\ N}}\put(9,0){\vector(-1,0){5}}\put(3,1){\sf{f}}\put(28,17){\sf{a}}\end{picture}$

Given,

• mass of the block,

$\longrightarrow\sf{m=70\ kg}$

• coefficient of friction,

$\longrightarrow\sf{\mu=0.5}$

The block is subjected to a force of 500 N which produces an acceleration a on the block in the same direction as that of the applied force.

From the fig.,

$\longrightarrow\sf{R=700\ N}$

• The frictional force acting on the block,

$\longrightarrow\sf{f=\mu R}\\\\\longrightarrow\sf{f=0.5\times 700}\\\\\longrightarrow\sf{f=350\ N}$

Then, the net force acting on the block,

$\longrightarrow\sf{70a=500-f}\\\\\longrightarrow\sf{70a=500-350}\\\\\longrightarrow\sf{70a=150}\\\\\longrightarrow\large\boxed{\boxed{\sf{a=}\ \bf{2.14\ m\ s^{-2}}}}$