Question

A block of mass 0.5kg has an initial velocity of 10 m/* down an inclined plane 30°, the coefficient of friction between the block and the inclined surface is 0.2.

Answer 1

Doward acceleration

a = gsin30°-mu×gxcis30°

=10×1/2-0.2×10×√3/2

=5-√3 m/*^2

So velocity after moving 10 m downward

v^2=u^2+2×a×10

=>v^2= 10^2+2×(5-√3)×10

=v=13m/* (approx)