Question

a force of 100 g wt is required to pull a body weighing 1 kg over ice . what is the coefficient of friction ? take g as 9.8 m/s^2​

Answer 1

hey...

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:-

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

Answer 2

Hey MaTe...!!

YouR AnSwEr is...!!

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

m = 1 kg = 1000 g

R = mg = 1000 × 980 dyne

f = µR

µ = f/R

µ = 100*980÷ 1000*980

Further solving....

we get

µ = 1/10

µ = 0.1

HOpe iT HelP YOu DEaR

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