A ball rolling on ice with a velocity of 5.6 m per second stops after travelling 8m find the coefficient.
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Hey mate!
Here's your answer!!
We know that,
v²-u²=2as
0-(5.6)²=2 × a × 8
a= - (1.96)
Coeff. of friction = fms × N
= ma × mg
=m²(a × g)
= m²(1.96 × 10)
Coefficient of friction = m²(19.6)
✌ ✌ ✌
#BE BRAINLY
Here's your answer!!
We know that,
v²-u²=2as
0-(5.6)²=2 × a × 8
a= - (1.96)
Coeff. of friction = fms × N
= ma × mg
=m²(a × g)
= m²(1.96 × 10)
Coefficient of friction = m²(19.6)
✌ ✌ ✌
#BE BRAINLY
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