Question

force of 147 N is required to just slide a block of weight 500N over a surface of ice calculate the coefficient of friction between block and ice​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}}$

• Given :

=> external force = 147 N

=> weight force = 500 N

=> Normal reaction = weight force = 500N

• To Find :

=> Co-efficient of friction

• Formula :

$\implies \: \boxed{ \pink{frictional \: force = \mu \times </strong><strong>N</strong><strong>}}$

=> Frictional force = External force

• Calculation :

$\implies \: f = \mu \times </strong><strong>N</strong><strong> = </strong><strong>F</strong><strong>ext \\ \\ \implies \: \mu \times 500 = 147 \\ \\ \implies \: \mu = \frac{147}{500} = 0.294 \\ \\ \huge{ \star} \: \boxed{ \boxed{ \blue{ \mu = 0.3}}}$

Answer 2

$\Huge{\underline{\underline{\mathfrak{ Solution \colon }}}}$

From the Question,

• Force acting on the block,F = 147 N

• Weight,W = 500 N

To finD

Coefficient of Friction

From Newton's Second Law,

$\sf \: F_{net} = Ma$

The block isn't still stationary,thus a = 0 m/s². Thus,the net force acting on the block is zero

Now,

$\sf{F - f = 0} \\ \\ \longmapsto \: \sf{F = \mu \: N}$

Here,

The normal reaction and weights are balanced here. Net force along Y axis is also zero. Thus,the normal reaction is also equal to 500 N.

$\longmapsto \ \sf \: 147 = \mu \: \times 500 \\ \\ \longmapsto \: \sf \mu \: = 0.29 \\ \\ \longmapsto \: \boxed{ \boxed{ \sf{ \mu = 0.30}}}$