A object has a mass of 6 kg.It is thrown in a plane surface.The friction of the surface is 0.2N.The initial velocity of the object is 20m/s and final velocity is 10m/s.it takes 2 s time to get the final velocity.Then what will be the applied force on the object?
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The resultant force acting down on the body along a slope
F=mg \sin\theta=2\:\rm{kg}\times 9.8\:\rm{N/kg}\times\sin 30^{\circ}=9.8\:\rm{N}F=mgsinθ=2kg×9.8N/kg×sin30
∘
=9.8N
The acceleration of the body
a=\frac{F-F_{\rm{fric}}}{m}=\frac{9.8\:\rm{N}-5\:\rm{N}}{2\:\rm{kg}}=2.4\:\rm{m/s^2}a=
m
F−F
fric
=
2kg
9.8N−5N
=2.4m/s
2
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