1.

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»

\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}}}


Discussion

No Comment Found

Related InterviewSolutions