1. Work, power and energy(2)
Solutions

2. W = mg = (50)(10) = 500N P.E. = mgh P.E. = (50)(10)(4) = 2000J
Q1. A pole-vaulter has a mass of 50kg. (a) What is her weight in newtons?(g = 10N/kg)
W = mg
= (50)(10) = 500N
(b) If she vaults to 4 m high, what is her gravitational potential energy?
P.E. = mgh
P.E. = (50)(10)(4)
= 2000J

3. (c) How much kinetic energy does she have just before reaching the ground?
P.E. at top = K.E. at bottom
K.E. just before reaching the ground – 2000J

4. Q2. A car of mass 1000kg is travelling at 30m/s
Q2. A car of mass 1000kg is travelling at 30m/s. (a) What is its kinetic energy?
K.E. = ½ mv2
=½ (1000)(30)2 = J
(b) If slows to 10m/s. What is its K.E. now?
K.E. = ½ mv2
=½ (1000)(10)2 = 50000J

5. Q2. (c) What is the change in kinetic energy?
Change in K.E. = Initial K.E. – Final K.E.
= – = J
(d) If it takes 80m to slow down by this amount, what is the average breaking force?
Work Done = Change in energy = Force x Distance
= Braking Force x 80
Average Breaking force =  80 = 5000N

6. Q3. A girl throws a ball upwards at a velocity of 10m/s
Q3. A girl throws a ball upwards at a velocity of 10m/s. How high does it go?(g = 10N/kg)
K.E. at bottom = P.E. at top
½ mv2 = mgh
½ (10)2 = 10h
h = 5m

7. Q4. An electric lamp is marked 100W
Q4. An electric lamp is marked 100W. How many joules of electrical energy are tranformed into heat and light, (a) during each second.
100W = 100J/s
100J of energy converted every second
(b) During a period of 100s.
No. of joules during 100s = 100 x 100
= 10000J

8. K.E.sprinter = ½ mv2 = ½ (60)(10)2 = 3000J
Q5. Calculate the kinetic energy of a sprinter of mass 60kg running at 10m/s.
K.E.sprinter = ½ mv2
= ½ (60)(10)2
= 3000J

9. Q6. A free-wheeling motor cyclist of mass (including her machine) 100kg is pushed from rest over a distance of 10m. If the push of 250N acts against a frictional force of 70N, calculate the kinetic energy and velocity when the push ends.
200N
m =100kg
70N
Resultant Force = = 180N to the right
Work Done = Force x Distance = Energy Changed
= 180 x 10 = 1800J (KE when push ends)

10. KE when the push ends is 1800J K.E. = ½ mv2 1800 = ½ (100)v2
Solution Q6 contd.
KE when the push ends is 1800J
K.E. = ½ mv2
1800 = ½ (100)v2
v = ((1800 x 2)  100)
= 6 m/s
Velocity when push ends = 6m/s

11. Q7. A grandfather clock uses a mass of 5kg to drive its mechanism
Q7. A grandfather clock uses a mass of 5kg to drive its mechanism. Calculate the gravitational potential energy stored when the mass is raised through its maximum height of 0.8m.
P.E. = mgh
= (5)(10)(0.8)
= 40J

12. Q8. (a) What is the velocity of an object of mass 1kg which has 200J of K.E.?
K.E. = ½ mv2
200 = ½ (1)v2
v2 = 400
v = 20m/s

13. (i) P.E. = mgh P.E. = (5)(10)(3) = 150J (ii) P.E. = mgh
Q8. (b) Calculate the p.e. of a 5kg mass when it is (i) 3m, (ii) 6m, above the ground. (g = 10N/kg)
(i) P.E. = mgh
P.E. = (5)(10)(3)
= 150J
(ii) P.E. = mgh
P.E. = (5)(10)(6)
= 300J

14. Q9. A 100g steel ball falls from a height of 1
Q9. A 100g steel ball falls from a height of 1.8m on to a plate and rebounds to a height of 1.25m. Find (a) the PE of the ball before the fall (g=10m/s2)
1.8m
1.25m
M = 100g = 0.1kg
PEtop = mgh
PE = (0.1)(10)(1.8)
= 1.8J

15. Q9. (b) the KE as it hits the plate,
1.8m
1.25m
M = 100g = 0.1kg
PEtop = KEbottom
KE = 1.8J

16. Q9. (d) the KE as it leaves the plate on the rebound,
1.8m
rebounds1.25m
M = 100g = 0.1kg
Petop = KEbottom
rebounds to 1.25m
Energy at bottom =mgh
= (0.1)(10)(1.25)
= 1.25J

17. Q9. (e) its velocity on rebound,
1.8m
rebounds1.25m
M = 100g = 0.1kg
½(0.1)v2 = 1.25
V = 5m/s

18. Loss in PE = Gain in KE Gain in KE = 1000J
Q10. A body of mass 5kg falls from rest and has a KE of 1000J just before touching the ground. Assuming there is no friction and using a value of 10m/s2 for the acceleration due to gravity, calculate (a) the loss of PE during the fall.
Loss in PE = Gain in KE
Gain in KE = 1000J

19. Q10. (b) the height from which the body has fallen.
PEtop = mgh
1000 = (5)(10)h
h = 20m