A particle of mass 200 g moves in a straight line such that the graph of the square of its velocity ( v^2 ) against displacement ( x ) is a straight line. The graph passes through the points (x=0, v^2=50) and (x=10, v^2=10) in SI units. If the magnitude of the retarding force acting on the particle is N 10⁻¹ N , the value of N is ________.

Step-by-step solution

The slope of the v^2 versus x graph is given by: Slope = v_f^2 - v_i^2 x_f - x_i = 10 - 50 10 - 0 = -4 m/s ^2 We know that v^2 = u^2 + 2ax . Differentiating with respect to x , we get: d(v^2) dx = 2v dv dx = 2a Therefore, 2a = -4 a = -2 m/s ^2 . The mass of the particle is m = 200 g = 0.2 kg . The retarding force is F = ma = 0.2 (-2) = -0.4 N . The magnitude of the force is 0.4 N = 4 10⁻¹ N . Hence, N = 4 . Answer: 4