An object of mass 10kg moving with a velocity of 5.6ms-1 on a horizontal rough surface comes to rest after traveling a distance of 16km, then the work done against friction is
02madhumathi1:
Candoina physics method
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a = u^2 / (2S)
= 5.6^2 / (2 * 16 * 10^3)
= 0.00098 m/s^2
W = F × S
= ma × S
= 10 kg × 0.00098 m/s^2 × 16 km × 10^3 m/km
= 156.8 joule
Work done against friction is 156.8 joule
= 5.6^2 / (2 * 16 * 10^3)
= 0.00098 m/s^2
W = F × S
= ma × S
= 10 kg × 0.00098 m/s^2 × 16 km × 10^3 m/km
= 156.8 joule
Work done against friction is 156.8 joule
I want physics method
Physics method? Sorry, I didn’t understood what you said.
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By equation v^2 =u^2 +2 we have, a=u^2 \2as a=5.6ms^-1\2*16*10^3metre a=0.00098ms^-2 w=F*S w=ma*s w=10*0.00098\16*10^3 w=156.8 joule
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