A child of weight 330N is at point X at the top of a slide. The slide is at the edge of a swimming pool, as shown in Fig. 3.1. X child, weight 330 N surface of slide 4.0 m surface of water Y 1.1 m water in swimming pool Fig. 3.1 (not to scale) The child moves from rest to the lowest point of the slide that is a vertical distance of 4.0 m below X. The child continues moving towards point Y which is at the end of the slide and a vertical distance of 1.1 m above the lowest point. The kinetic energy of the child at Y is 540 J.
- (a)Calculate the difference in the gravitational potential energy of the child at points X and Y.[2]
- (b)An average frictional force of 52N acts on the child when moving from X to Y. By considering changes of energy, determine the distance moved by the child from X to Y.[2]
- (c)(i)The child leaves the slide at point Y with a velocity that is at an angle of 41° to the horizontal. The path of the child through the air is shown in Fig. 3.2. velocity Z Y 41° path of child slide surface of water water in swimming pool Fig. 3.2 (not to scale) Point Z is the highest point on the path of the child through the air. Assume that air resistance is negligible. Calculate the speed of the child at: point Y[2]
- (c)(ii)Calculate the speed of the child at: point Z.[2]
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- [C1] (Δ)E = mg(Δ)h or W(Δ)h; = 330 × (4.0 – 1.1)
- [A1] = 960 J
- [C1] (work =) 960 – 540; (= 420 J)
- [A1] distance moved = (960 – 540) / 52; = 8.1 m
- [C1] E = ½mv²; 540 = ½ × (330/9.81) × v²
- [A1] v = 5.7 m s⁻¹
- [C1] speed = horizontal component of velocity; = 5.7 × cos 41°
- [A1] = 4.3 m s⁻¹
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