Physical significance of Physical Constants in Laws of Physics

Beyond Einstein! Within Einstein's Theory

First Year Physics Notes chap 4, Work and Energy

4.7: Conservation of Energy

Energy can neither be created nor destroyed but can transform from one to another such that toatal energy remains constant

Total Energy = PE + KE

SHORT QUESTIONS AND ANSWERS FROM EXERCISE

4.1: A person holds a bag of groceries, while standing still, talking to a friend. A car is stationary with its engine running. From stand point of  work, how are these two situation similar?

Ans: In both cases no displacement is present so work done is zero in both case

4.2: Calculate the work done in kilojoules in lifting a mass of 10 kg at a steady velocity through vertical height of 10 m.

Ans: mass m = 10 kg,      h = 10 m,    g = 9.8ms^-2,  w = ?

W = mgh

W = 10 ´ 9.8 ´10 J

W = 980 J = 0.98 kJ

4.4: In which case more work is done?  (i) When a 50 kg bag of books is lifted through 50 cm or  (ii) when a 50 kg crate is pushed through 2m across the floor with a force of 50N?

Ans: (i) w = mgh

W = 50 ´ 9.8 ´ ½ J

W = 25 ´ 98/10 J = 5 ´49J

W = 245 J

(ii) W = F´d

W = 50 ´2 J

W = 100 J

In case (i), more work is done

4.5: An object has 1 J of Potential energy. Explain what does it mean?

Ans: We know that

PE = mgh

The object has weight of IN at a height of  1 m or possesses 1 J  potential energy

4.6: A ball of mass m is held at a height h₁ above the table. The table top is at a height h₂ from the floor. One student says that the ball has PE = mgh₁ but another says it is mg(h₁ +h₂). Who is correct?

Ans: Both students are correct according to their observation

First student is considering the PE from the top of the table while the other is looking from the floor to the top

4.7: When a rocket re-enters the atmosphere, its nose cone becomes very hot. Where does this heat come from?

Ans: When the rocket re-enters in the atmosphere, its velocity increases rapidly so due to high kinetic enrgy and high friction, its nose cone becomes very hot.

4.8: What sort of energy is in the following?

(i) Compressed spring  (ii) Water in a high Dam  (ii) A moving car

Ans: (i) Elastic PE   (ii) Gravitational PE (ii) Kinetic Energy

4.9: A girl drops a cup from a certain height, which breakes into pieces. What energy changes are involved?

Ans: Potential energy of the cup is converted into KE, heat and sound

4.10: A boy uses a catapult to throw a stone which accidently smashes a green house window. List the possible energy changes?

Ans: The elastic potential energy of the catapult is converted into kinetic energy of stone, heat and sound produced

NUMERICAL PROBLEMS OF EXERCISE

4.1: A man pushes a lawn mover with a 40 N force directed at an angle of 20 downward from horizontal . Find the work done by the man as he cuts a strip of  grass 20 m long?

Data:

 F = 40 N,      d = 20 m,           q = 20,    w = ?

Solution:

W = Fdcosq

W = 40 ´ 20´cos20

W = 750 J

4.2: A rain drop of mass 3.5  10^-5 kg falls vertically at a constant speed under the influence of force of gravity and friction.  In falling through 100 m, how much work is done by (i) gravity   (ii) friction

Solution:

W = mgh

W = 3.5  10^-5 ´ 9.8 ´ 100 J

W = 0.0328 J

W = - 0.0328 J

4.3: Ten blocks, each 6cm thick and mass 1.5 kg, lie flat on a table. How much work is required to stack them one on the top of another?

Data:

Mass m = 1.5 kg,   height h = 6cm = 0.06 m, work done = ?

Solution:

W = mgh

Work done to place 2^nd brick on ist, 3^rd on 2^nd and finally 10^th on 9^th

W_2®1 = mgh₁ = 1.5´9.8 ´1( 0.06) J

W_3®2 = mgh₂ = 1.5´9.8 ´2( 0.06) J

W_4®3 = mgh₃ = 1.5´9.8 ´3( 0.06) J

.

.

.

W_10®9 = mgh₉ = 1.5´9.8 ´9( 0.06) J

W_total = mgh = 1.5´9.8 ´[1 + 2 + 3 + 4 + … + 9]( 0.06) J

W_total = mgh = 1.5´9.8 ´45( 0.06) J

W_total = 40 J

4.4: A car of 800 kg travalling at 54 kmh^-1 is brought to rest in 60 m. Find the average retarding force on the car. What has happened to original kinetic energy?

Data;

M = 800 kg,   v_i = 54km/h = 15 m/s,    v_f = 0,    s= 60 m,    F = ?

Solution:

Using work energy principle

Fs = ½ m(v_f² -  v_i²)

Fs = ½ m(0 -  v_i²)

F = -½ mv_i²/s

F = -½ ´ 800 ´225/60 N

F = -1500 N

4.5:  A 1000 kg automobile atb the top of an incline 10 m high 100 m long is released and rolls down the hill. What is its speed at the bottom of the incline if the average retarding force due to friction is 480 N?

Data:

M = 1000 kg,     v_i = 0,   v_f = ?,   s = 100 m,    h = 10 m,   F = 480 N

Solution:

W = DKE

Fs = ½ m(v_f² -  v_i²)

Fs = ½ m(v_f² -  0)

v_f²  = 2Fs/m

v_f²  = 2´480´100/1000

v_f   = 9.7 ms^-1

4.6: 100 m³ water is pumped from a reservoir into a tank, 10 m higher than the reservoir, in 20 minutes. If density of water is 1000 kgm^-3.  Find

(a) Increase in PE           (b) power delivered

Data:

Volume v = 100 m³,    height h = 10 m,   time t = 20 min = 1200 s,   PE = ? P = ?

Solution:

PE = mgh                         m = r´ V = 1000´100 Kg

PE = 1000000´ 9.8 J

PE = 9.8 ´10⁶ J

Power = mgh/t

Power = 1000000´ 9.8/1200

Power = 8.2 ´10³ watts

4.7: A force of 400 N is required to overcome road friction and air resistance in propelling an automobile at 80 kmh^-1. What power (kw) must be engine develop?

Data:

 F = 400 N,    v = 80 kmh^-1 = 22.22ms^-1,   Power P = ?

Solution:

P = F.v

P = 400 ´ 22.22 watts

P = 8900 watts

P = 8.9 kw

4.8: How large a force is required to accelerate an electron of mass 9.1´10^-31 kg from rest to a speed 2´10⁷ ms^-1 through a distance of 5 cm?

Data:

m =  9.1´10^-31 kg,      v_i = 0,        v_f = 2´10⁷ ms^-1,    d = 5 cm = 0.05m,    F= ?

Solution:

W = DKE

Fs = ½ m(v_f² -  v_i²)

Fs = ½ m(v_f² -  0)

F = ½ mv_f² /s

F = ½ ´9.1´10^-31 ´2´10⁷ ´2´10⁷  /0.05 N

F = 3.6´10^-15  N

4.9: A diver weighing 750 N dives from a board of 10 m above the surface of the pool of water. Use the conservation of mechanical energy to find his speed at a point 5 m above the water surface, neglecting the air resistance/

Data:

W = mg = 750 N,     h₁ = 5m,   h₂ = 10 m,   v₁ = 0,   v₂ = ?

Solution:

Gain in KE = loss in PE

½ m(v₂² -  v₁²) = mg(h₂ -  h₁)

½ m(v₂² -  0) = mg(h₂ -  h₁)

v₂²  = 2 g(h₂ -  h₁)

v₂²  = 2 ´9.8(10 -  5)

v₂²  = 2 ´9.8 ´ 5 = 9.8 ´ 10 = 98

v₂  =  Ö98

v₂  =  9.9 ms^-1

4.10: A child starts from rest at the top of a slide of height 4 m (a) What is the speed at the bottom if the slide is frictionless (b) If he reaches the bottom, with the speed of 6 ms^-1. What percentage of this total energy  at the top of the slide is lost as a result of friction?

Data:

h₁ = 0m,   h₂ = 4 m,   v₁ = 0,   v₂ = v = ?

Solution:

Gain in KE = loss in PE

½ m(v₂² -  v₁²) = mg(h₂ -  h₁)

½ m(v₂² -  0) = mg(h₂ -  0)

v₂²  = 8g

v₂  = Ö8´9.8

v₂  = 8.85 ms^-1

Special Questions

Q: How can we get relation between energy and work in a single step?

Ans: If we multiply 3^rd equation of motion by mass m then we get work energy relation in a single step

2as = v_f² - v_i²

2mas = mv_f² - mv_i²

Fs = ½ mv_f² - ½ mv_i²

W = D( ½ mv_f)

Q: Is there any formula for work other than W = F.d?

Ans: Yes, it exists

W = F.d

Writing F = P/t

W = P/t.d

W = P.d/t                               but d/t = v

W = P.v

Q: Can we get relation for absolute gravitational potential energy from Newton’s law of gravitation?

Ans: We know that

F = G mM//R²   

Multyplying by R on both sides

FR = - GmM/R

U_g = - GmM/R                          FR = U_g

Negative sign is taken due to work done against gravity