Question no 4th & 5 th plze plze plze
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Force acting along the inclination is = mgsinΘ - μmgcosΘ
so acceleration a =(mgsinΘ - μmgcosΘ)/m
=gsinΘ - μgcosΘ
S =ut +1/2 a t2
8=0+1/2 (gsinΘ - μgcosΘ)*4
or 4=10*1/2 - μ*10*√3/2
μ=1/5√3
so acceleration a =(mgsinΘ - μmgcosΘ)/m
=gsinΘ - μgcosΘ
S =ut +1/2 a t2
8=0+1/2 (gsinΘ - μgcosΘ)*4
or 4=10*1/2 - μ*10*√3/2
μ=1/5√3
Anonymous:
But ans given is 5/√3
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