# Work, power and energy(2)

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Work, power and energy(2)
Solutions

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

(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

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

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

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

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

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

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)

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

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

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

(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

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

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

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

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

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

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