Question

A force of 98n is just able to move a block of mass 20 kg on a rough horizontal surface .Calculate the coefficent of friction

Answer 1

$\huge{\boxed{\text{Answer :-}}}$

♦ As provided that

→ Force applied to move = 98 N

→ Mass of Block = 20 kg

♦ Before solving we must know

• What equation for force ?

→ F = Mass × acceleration

>> F = ma

• What is equation for Kinetic friction ?

→ $F_k = \mu N$

• How to find $N$ ( Normal force) ?

→ Normal force is a contact force applied by the surface to the object .

N = MgCos∅

♦ Now solving the Question :-

>> As according to the Questions , it says that the force of 98 N was just able to move the block .

>> Then $F_k = Force \: applied$

♦ So

>> Normal force

= mg ( as cos∅ = 0° )

= 20 × 9.8

= 196 N

>>As $F_k = 98 \:\textbf{N}$

Then

$98 = \mu N$

$\implies 98 = \mu 196$

$\implies \mu = \dfrac{98}{196}$

$\implies \mu = \dfrac{1}{2}$

♦ So the value of coefficient of friction = $\bold{\dfrac{1}{2}}$

Answer 2

Hello ☺

___________________rrjack29 ❤ here

___________Answer ⤵⤵⤵

Given,

↪ Force applied to move = 98 N
↪ Mass of Block = 20 kg 

F = Mass × acceleration 

　◾　F = ma 


= mg

= 20 × 9.8

= 196 N

As, Fk ​=98N 

Then,

μ = 1/2

♦ So the value of coefficient of friction = 1/2