Question

Show all your steps and draw diagrams where necessary. Add explanatory notes where necessary. 1. A 20-kg child swings back and forth on a swing such that her height h in m from the ground as a function of x is described by: h(x) = 0.5 + 0.45x2 for -1.2 m  x  1.2 m (a) What is the position of the child when x = 0? (b) Show graphically how the potential and kinetic energies of the child vary with x. (c) What is the maximum potential energy of the child? (d) At what point does the child have maximum speed? (e) Use the law of conservation of energy to calculate the maximum speed of the child. 2. A spring of spring constant k = 340 Nm-1 is used to launch a 1.5-kg block along a horizontal surface by compressing the spring by a distance of 18 cm. If the coefficient of sliding friction between the block and the surface is 0.25, how far does the block slide? 3. An object of mass 4 kg is initially at rest on a frictionless ice rink. It suddenly explodes into three pieces (I have no idea why it happened!). One chunk of mass 1 kg slides across the ice with a velocity 1.2 ms-1 i and another chunk of mass 2kg slides across the ice with a velocity 0.8 m.s-1 j. Determine the velocity of the third chunk in magnitude angle form

Answer 1

( a)

Position of the child when x =0

h(0) = 0.5 + 0.45( 0 )² = 0.5 m

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(b)

Graph description:

Potential energy curve is symmetrical U shape. Here, highest points are at where x=-1.2 and x=1.2 and the lowest point of the graph is higher than zero, and it should be at where x=0.

Kinetic energy iscurve is inverted U shape, where the highest point is at x=0. The lowest points of the graph are at where x=-1.2 and x=1.2 and they touch at zero.

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(c) At extremme postion, the potential energy should be maximum, that is at x = 1.2. In this case,

h(1.2) = 0.5 + 0.45( 1.2 )² = 1.148 m

Use potential energy equation,

PE = mgh = ( 20 kg ) ( 9.8 m/s²) ( 1.148 m ) = 225 J

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(d)

The child has maximum speed at equilibrium position,

The equilibrium position on his track is at x = 0.

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(e)

The maximum potential energy at extremme positoin is converted into kinetic maxium when it is at x =0

Thus, the maximum kinetic energy at equilibrium is   225 J

Use the equation for kinetic enregy

K E = 0.5 m v² = 225 J

0.5 ( 20 kg ) v² = 225 J

Solve the equation for v

v = 4.7 m/s