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A horizontal force of 1200 gf is applied to a 1200 g block , which rests on a horizontal surface . If the coefficient of friction is 0.2, find the acceleration produced in the block.

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4 years ago

```							1 gf = 0.0098 N Then,force applied on block in N is1200×0.0098=11.76NSo ,and friction is R×0.2= 1.2×10×0.2=2.4NSo,acc. Of the block is = net force/net mass.Therefore,net force=11.76-2.4=9.36NAnd acc. =9.36/1.2= 7.8m/sec.square
```
4 years ago
```							Here, mass,m=1200g=1.2kgApplied force,Fa=1200gf=1200× .0098N=11.76NNormal Reaction,R= 1.2kg × 9.8=11.76NSo, Frictional flrce, Ff=R×0.2= 11.76×0.2=2.35NThus, Net force, F= Fa-Ff=11.76-2.35 N=9.41NThus, Acceleration= F/m=9.41/1.2=941/120=7.841m/sec²
```
3 years ago
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