work, power and energy(2)

AQUINAS DIOCESAN GRAMMAR

Work, power and energy(2)

Solutions


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.(a) What is its kinetic energy?

K.E. = ½ mv2

=½ (1000)(30)2 = 450000J

(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.

= 450000 – 50000 = 400000J

(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

400000 = Braking Force x 80

Average Breaking force = 400000 80 = 5000N


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 = 10hh = 5m


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


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.

m =100kg 200N70N

Resultant Force = 250 - 70 = 180N to the right

Work Done = Force x Distance = Energy Changed

= 180 x 10 = 1800J (KE when push ends)


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. 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


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 1.8m 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 = 0.1kgPetop = KEbottom

rebounds to 1.25mEnergy 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


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

work, power and energy(2)

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