Question

A force of 100 g wt. is required to pull a body weighing 1 kg over ice. What is the coefficient of friction? [g = 9.8 m/s²]

Answer 1

$\mathfrak{\large {\underline {\underline{Answer \: : - }}}}$

∴ The coefficient of friction is 0.1

$\mathfrak { \large{\underline {\underline{Explanation \: : - }}}}$

Given :

f = 100 g wt = 100 × 980 dyne [∵ 1 g wt = 980 dyne]

m = 1 kg = 1000 g

R = mg = 1000 × 980 dyne

We know that :

$\bigstar \: \: \: \: \sf f = \mu </strong><strong>R</strong><strong>\:$

$\huge \implies {\boxed{ \sf { \mu = \frac{f}{</strong><strong>R</strong><strong>} }}} \:$

Solution :

$\sf \frac{100 \times 980}{1000 \times 980} = \frac{1}{10} = \sf\boxed{0.1}$

$\rule{300}{1.5}$

∴ The coefficient of friction is 0.1

Answer 2

Step-by-step explanation:

f = 100 g wt =

100 × 980 dyne

[∵ 1 g wt = 980 dyne]

m = 1 kg = 1000 g

R = mg = 1000×980dyne

Solution:

980*100/980*1000 = 0.1