Answers to Worksheet 6

pg 46-47

4.
Total energy at top = total energy at bottom + energy lost to surroundings
GPE + KE = GPE + KE + Energy lost
mgh + 0  =  0 + ½ x mv² + energy lost      [not moving at top=> no ke; no gpe at bottom]
2 kg x 10 x 6 m = ½ x 2 kg x 10² + energy lost

Energy lost = 20 J

5.
Since walking up a stairs, thus work is done against gravity.

Work done = F x d = Weight x height
                              = (80 kg x 10) x 4 m= 3200 J
Power = Work Done  t = 3200 J ÷ 16 s = 200 W

6.
Total Work Done at start = F x d = 8 N x 10 m = 80 J [input]

Work done against friction = F x d = 6 N x 10 m = 60 J [output]

Work done to increase KE = 80 - 60 = 20 J [output]

*energy is conserved as total energy input = total energy output

pg 50-51

5.
a. Power = Work Done ÷ t = 1200 J ÷ 10 s = 120 W
b. GPE = mgh = 50 kg x 10 x 2 m = 1000 J
c. Some energy lost as / converted to heat due to friction between student and the stairs.

6.
a. GPE = mgh = 20000 kg x 10 x 10 m = 2000000 J
b. distance = Work done ÷ Force = 345000 J ÷ 50000 N = 6.9 m = 7 m (1sf)

7.
a. Work done = F x d = 50 N x 3 m = 150 J
b. Work done = F x d = Weight x height = (10 kg x 10) x (3 x 9)    [no of floors between 1 and 10 is 9]
                                                              = 2700 J

8.
a. GPE is converted to electrical energy
    Power = energy converted ÷ t
               = mgh ÷ t
               = (m ÷ t) x g x h = 4000 kg/s x 10 x 600 m = 24000000 W
b. if 90% efficient, electrical energy = 90% x answer in (a) = 21600000W

pg 52
1a.
i.  Work done = F x d = 26 N x 4 m = 104 J  [4 m is in direction of force]
ii. Gain in KE = ½ x mv² = ½ x 2 kg x 8² = 64 J
iii. Gain in GPE = mgh =  2 kg x 10 x (4 sin 30°) = 40 J
[sin 30° = opp ÷ hypo = height ÷ 4 => height = 4 sin 30°]

1b.   i = ii + iii. Energy is conserved. Work done [input] is converted to KE and GPE [output]