Question

An 850 kg car can roll up and down a ramp at an angle of theta = 45.0 degree to the ground. It is connected, by a massless tow rope that runs over a frictionless pulley, to a heavy spring of spring constant k = 35.0 kN/m, as shown in the figure to the right. The car handbrake is released when the car is at rest, with the spring in its unstretched state. Assume the rolling motion of the car is frictionless and that the mass of the wheels is negligible compared to the main body. (i) What is the speed of the car When it has move 5.00 cm down the ramp (ii) How far down the ramp from its point of release does the car roll before momentarily stopping?

Answer 1

Weight of car, m = 850 kg spring constant, k = 35 kN/m Force exerter by the car through the ramp = mg sin theta Force produced by the spring when it was stretched upto 5 cm = 0.05 m Therefore F_sp = k middot x = 35 times 0.05 = 1.75 kN Therefore Average Force produced by the spring when it is stretching upto 0.05 cm = 0 + 1.75 /2 = 0.875 kN = 875 N If the acceleration of the car is alpha (let), then as from the given condition mg sin theta - 875 = m alpha alpha = mg sin theta - 875 /m = (850 times 9.8 times sin 45 degree) - 875 /850 alpha = 5.9 m/s^2 Therefore As per law of kinematics, v^2 = u^2 + 2 alpha s [u = Initial velocity v = final velocity s = Displacement /alpha = acceleration] v^2 = (0)^2 + (2 times 5.9 times 0.05)