RGS 1998 P2 Prelims Q9
A block of mass 1 kg slides down a plane with an angle of inclination of 30° as shown in the first figure below. The speed-time graph of the block is shown in the 2nd figure. PQ denotes the motion of the block in portion XY while QR denotes that in portion YZ. (Portions XY and YZ are made of different materials).
(a)
(i) Indicate the forces acting on the block in the first figure.
(ii) The weight of the block is the resultant force of two force components; one is parallel to the slope of the inclined plane and the other is perpendicular to the slope of the inclined plane. By means of a scale diagram, determine the magnitude of these two force components.
(iii) Hence, determine the frictional force acting on the block in the portion XY of the inclined plane.
(b) Determine from the graph in the 2nd figure
(i) the acceleration of
(ii) the distance travelled by
the block in the portion YZ of the inclined plane.
(c) If the block is projected upwards from Z along the inclined plane with a sufficiently large initial speed (so that the block passes through Y), sketch the speed-time graph of the motion up the plane.
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Answer:
(a)
(i)
(ii) Draw to scale with 1 cm to 1 N, with the angles shown below
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Force component perpendicular to surface = 8.66 N
Force component parallel to surface = 5 N
(iii) For portion XY, there's zero acceleration as seen from the speed-time graph.
Hence, net force along surface = 0
Thus, frictional force = 5 N
(b)
(i) From graph,
initial speed = 0.75 m/s
final speed = 0 m/s
thus, acceleration = (0 - 0.75) / 2 s
acceleration = -0.375 m/s2
(ii) Distance travelled = area under graph
Distance travelled = ½ (0.75) (2) = 0.75 m
(c) It can be seen from part (a) that the friction for portion YZ is greater than for portion XY
Thus, together with gravitational acceleration along the slope, the deceleration for portion YZ will be greater than for portion XY
AB represents for portion YZ
BC represents for portion XY
