A 40 kg block accelerates by force of 200N. acceleration of the block

1.	A 40 kg block accelerates by force of 200N. acceleration of the block is 3 m/s^2 determine the magnitude of friction force fg experienced by the block
Answer» Given : Mass of block = 40kg Applied force = 200N Acceleration of the block = 3m/s² To Find : Magnitude of friction force EXPERIENCED by the block. Solution : ❒ As per newton's second law of motion, force is measured as the product of mass and acceleration. Mathematically, F = m × a F denotes force m denotes mass a denotes acceleration We can use this formula only when friction is absent. In this QUESTION, friction force OPPOSES the forward motion. Therefore the equation will be, F - f = ma where f denotes friction force $\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{40\:kg}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(<klux>3</klux>,0){19}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\put(32,6){\sf 200\:N}\end{picture}$ By substituting the given values; ➙ F - f = ma ➙ 200 - f = (40)(3) ➙ 200 - f = 120 ➙ f = 200 - 120 ➙ f = 80 N

A 40 kg block accelerates by force of 200N. acceleration of the block is 3 m/s^2 determine the magnitude of friction force fg experienced by the block

Answer»

Given :

Mass of block = 40kg

Applied force = 200N

Acceleration of the block = 3m/s²

To Find :

Magnitude of friction force EXPERIENCED by the block.

Solution :

❒ As per newton's second law of motion, force is measured as the product of mass and acceleration.

Mathematically, F = m × a

F denotes force
m denotes mass
a denotes acceleration

We can use this formula only when friction is absent. In this QUESTION, friction force OPPOSES the forward motion.

Therefore the equation will be,

F - f = ma

where f denotes friction force

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\framebox(15,15){\sf{40\:kg}}}\put(-20,-0.1){\line(1,0){55}}\multiput(-19,0)(<klux>3</klux>,0){19}{\qbezier(0,0)(-1,-1)(-2,-2)}\put(15,7.5){\vector(1,0){15}}\put(0,0.2){\vector(-1,0){15}}\put(-17,1.3){\sf{f}}\put(32,6){\sf 200\:N}\end{picture}$

By substituting the given values;

➙ F - f = ma

➙ 200 - f = (40)(3)

➙ 200 - f = 120

➙ f = 200 - 120

➙ f = 80 N