A bar of mass m resting on a smooth horizontal plane starts moving due to force | F → | = m g 9 . The magnitude of the force remains constant with time. The force vector makes an angle θ with the horizontal which varies with the distance covered as θ = Cx . If the constant C = 10     degree   meter   , then the speed of the bar, when θ becomes equal to 30 ° for the first

Step-by-step solution

Given, The magnitude of a force, | F → | = m g 9 and Force vector makes a varying angle θ with the horizontal as θ = C x , where, C = 10   degree   meter   . Thus, Horizontal component of force, F x = F   cos   θ = F   cos   C x . Applying work energy theorem, W net = ∆ K . E . ⇒ ∫ 0 x F x d x = 1 2 m v 2 (when θ becomes 30 ° , x = θ C = 30 ° 10 °   m - 1 = 3   m ) ⇒     ∫ 0 x m g 9 &#1