In a children park an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6-E1). Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g=10 m/s².Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
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Solution :
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let m be the mass of the child.
R-mg Cos 45 °=0
R=mgcos 45°=mg /√2
Net force acting on te body due to which it slides down.
=mg Sin 45 °- μR
=mg Sin 45°- μmg Cos 45°
=mx10 x 1/√2-0.6 x m x10 x 1/ √2
=m(5√2 - 3√2)
= m2√2
acceleration = F/m
a=m2√2/ m= 2√2 m/s2
***************************************************
let m be the mass of the child.
R-mg Cos 45 °=0
R=mgcos 45°=mg /√2
Net force acting on te body due to which it slides down.
=mg Sin 45 °- μR
=mg Sin 45°- μmg Cos 45°
=mx10 x 1/√2-0.6 x m x10 x 1/ √2
=m(5√2 - 3√2)
= m2√2
acceleration = F/m
a=m2√2/ m= 2√2 m/s2
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