Question

A trolley of mass 2-0 kg is catapulted up a slope. The slope is at an angle of 20 to the horizontal as shown in the diagram below. The speed of the trolley when it loses contact with the catapult is 3.0 ms1 slope trolley catapult 20° The size of the force of friction acting on the trolley as it moves up the slope is 1.3 N (a) ) Calculate the component of the weight of the trolley acting parallel to the slope (ii) Draw a diagram to show the forces acting on the trolley as it moves up the slope and is no longer in contact with the catapult. Show only forces or components of forces acting parallel to the slope. Name the forces (iii) Show that, as the trolley moves up the slope, it has a deceleration of magnitude 4.0 ms (iv) Calculate the time taken for the trolley to reach its furthest point up the slope (v) Calculate the maximum distance the trolley travels along the slope The trolley now moves back down the slope. (b)(i) Draw a diagram to show the forces acting on the trolley as it moves down the slope Show only forces or components of forces acting parallel to the slope. Name the orceS -2 (ii) The magnitude of the deceleration of the trolley is 4.0 ms as it moves up the slope Explain why the magnitude of the acceleration is not 4.0 ms when the trolley moves down the slope

Answer 1

Force f = friction force N = normal reaction mg sin 20 degree & mg cos 20 degree = parallel and perpendicular components of weight a, v = acceleration and velocity at any time. (a) i) parallel component of weight we = mg sin 20 degree w_p = 2 times 9.81 times sin 20 degree w_p = 6.71 N ii) see above iii) here, the horizontal component of forces are (along the plane) F_net = f + mg sin 20 degree m times a = 1.3 + 6.7 2 times a = 8 a = 4 m/s^2 hence proved \r\n iv) we know that at highest point velocity should be zero v = u + at 0 = 3 + (-4) times t deacceration is acting oppicate to velocity t = 3/4 t = 0.75 sec v) Again, using vee^2 = u^2 + 2 as 0^2 = 3^2 + 2 (- 4) times s 9 = 85 s = 1.125m