A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated. (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Given,
(a)Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5 m
Number of times the weight is lifted, n = 1000
Therefore,
Work done against gravitational force,
W = n(mgh)
= 1000 × 10 × 9.8 × 0.5
= 49 kJ
(b)Energy equivalent of 1 kg of fat = 3.8 × 107 J
Efficiency rate = 20%
Mechanical energy supplied by the person’s body,
E = (20/100) × 3.8 × 107 J
= (1/5) × 3.8 × 107 J
Equivalent mass of fat lost by the dieter,
m = [ 1 / (1/5) × 3.8 × 107]× 49 × 103
= (245 / 3.8) × 10-4
= 6.45 × 10-3 kg
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A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Given,
Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, ub = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, uB = 0
Final speed of the system of the bullet and the block = ν
Using the law of conservation of momentum,
mub + MuB = (m + M) v
0.012 × 70 + 0.4 × 0 = (0.012 + 0.4) v
∴ v = 0.84 / 0.412 = 2.04 m/s
For the system of the bullet and the wooden block,
Mass of the system, m' = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system,
Potential energy at the highest point = Kinetic energy at the lowest point
m'gh = m'v2
∴ h =
= ×
= 0.2123 m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet – Kinetic energy of the system
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3.
What is the electrical power produced?
Given,
Area of the circle swept by the windmill = A
Velocity of the wind = v
Density of air = ρ (a) Volume of the wind flowing through the windmill per sec = Av
Mass of the wind flowing through the windmill per sec = ρAv
Mass m, of the wind flowing through the windmill in time t = ρAvt (b) Kinetic energy of air = mv2
= (ρAvt)v2
= (1/2)ρAv3t (c) Area of the circle swept by the windmill, A = 30 m2 Velocity of the wind = v = 36 km/h
A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
(a)
Power used by the family, P = 8 kW
= 8 × 103 W
Solar energy received per square metre = 200 W
Efficiency of conversion from solar to electricity energy = 20 %
Area required to generate the desired electricity = A
We have
8 × 103 = 20% × (A × 200)
= (20 /100) × A × 200
∴ A = 8 × 103 / 40 = 200 m2
(b)
The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof of a building having dimensions 14 m × 14 m. (≈ √200).
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A body of mass 0.5 kg travels in a straight line with velocity v = ax3/2 where a = 5 m -1/2 s-1. What is the work done by the net force during its displacement from x = 0 to x= 2 m?
m'v2
× 
mu2 - 
m'v2 
× 0.012 × (70)2 - 
× 0.412 × (2.04)2
mv2 
(ρAvt)v2 
) × Kinetic energy of air
ρ A v3t 
ρ A v3t / t
ρ A v3
× 1.2 × 30 × (10)3
) m (v2 - u2) 
) × 0.5 [ (10√2)2 - 02]
) × 0.5 × 10 × 10 × 2
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