Question

A force of 98 newton is required to pull a body of
mass 1x10^2 kg over the surface of the ice then
find the co-efficient of friction (g = 9.8 ms-)​

Answer 1

Answer:0.1

Explanation:

f=u(N).....(1)   [f=limiting friction(98newton) , u=co-efficient of friction, N=normal equilibrium reaction]

N=mg.......

=> N=10^2(9.8)

=>N=980 newton

therefore.., substituting values in equation (1)

u=1/10 or u= 0.1

Answer 2

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

A force of 98 newton is required to pull a body of  mass 1 x 10² Kg over the

surface of the ice

( g = 9.8 m/s² )

$\checkmark$ To Find:

Coefficient of friction ( μ )

$\checkmark$ Solution:

We know that,

$\red{\boxed{\sf f=\mu N}}$

Here,

• f = Frictional Force
• μ = Coefficient of friction
• N = Normal Force

We also know that,

$\green{\boxed{\sf N=mg}}$

Here,

• N = Normal Force
• m = Mass
• g = 9.8 m/s²

According to the Question,

We are asked to find the Coefficient of friction ( μ )

From the above given formulas,

Equating the Normal Forces (N)

Since,

$\sf \longrightarrow N=mg$

We get,

$\sf \implies f=\mu (mg)$

Hence,

$\blue{\sf \implies \mu=\dfrac{f}{mg}}$

Given that,

• f = 98 N
• m = 1 x 10² Kg
• g = 9.8 m/s²

Hence,

Substituting the values,

We get,

$\red{\sf \implies \mu=\dfrac{98}{1 \times 10^{2} \times 9.8}}$

$\green{\sf \implies \mu=\dfrac{98}{10^{2} \times 9.8}}$

$\blue{\sf \implies \mu=\dfrac{98}{980}}$

$\bold{\pink{\sf \implies \mu=0.1}}$

Therefore,

$\checkmark$ Coefficient of friction = 0.1