Y12 Questions on Energy and work

I have put in ^ to indicate a superscript (power)

1. (a) P.E. at top = 80 × 9.8(1) × 150 = 118 000 (J) (1)


K.E. at bottom and at top = 0 (1)

Elastic P.E. at top = 0, at bottom = P.E. at top for ecf = 118 000 J (1) 3

(b) 24 N m–1 × 100 m = 2400 N 1

(c) elastic P.E. is area under F-x graph (1)

graph is a straight line so energy is area of triangle (1)

elastic P.E. = ½ × kx × x = (½kx2) (1) 2

(d) loss of P.E. = 100 × 9.8(1) × 150 = 147 000 J (1)

gain of elastic P.E. = ½ × 26.7 × 1052 = 147 000 J (1) 2

(e) idea that a given (unit) extension for a shorter rope requires a greater force 1

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2. (a) (i) speed = d / t C1

= 24 / 55

= 0.436 (m s–1) allow 0.44 A1

do not allow one sf

(ii) kinetic energy = ½ m v2 C1

= 0.5 x 20 x (0.436)^2

= 1.9 (J) note ecf from (a)(i) A1

(iii) potential energy = mg h C1

= 20 x 9.8 x 4

= 784 (J) A1

penalise the use of g = 10

(b) (i) power = energy / time or work done / time C1

= (15 x 784) / 55

note ecf from (a)(iii)

= 214 (W) A1

(ii) needs to supply children with kinetic energy B1

air resistance B1

friction in the bearings of the rollers / belt B1

total mass of children gives an average mass of greater than 20 kg B1

Max B2

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