Class 11 - Unit # 05 : Work, Energy & Power - Solved Numericals
Physics Numerical Sheet: Work, Energy & Power
1. A man pulls a trolley through a distance of 10 m by applying a force of 50 N which makes an angle of 60° with the horizontal. Calculate the work done by the man?
Data:
Distance (d) = 10 m
Force (F) = 50 N
Angle (θ) = 60°
Work Done (W) = ?
Solution:
According to the definition of work done:
W = F × d × cos(θ)
Putting the given values:
W = 50 × 10 × cos(60°)
W = 500 × 0.5
W = 250 J
RESULT: Work done by the man is 250 joules.
2. A 100 kg man runs up a long flight of stairs in 9.8 second. The vertical height of the stair is 10 m. Calculate its power?
Data:
Mass (m) = 100 kg
Time (t) = 9.8 s
Height (h) = 10 m
Acceleration due to gravity (g) = 9.8 m/s²
Power (P) = ?
Solution:
According to the definition of Power:
P = W / t ---(i)
According to the definition of work done against gravity:
W = m × g × h
W = 100 × 9.8 × 10
W = 9800 J
Putting values in eq(i):
P = 9800 / 9.8
P = 1000 W
RESULT: The power of man is 1000 W.
3. When an object is thrown upwards. It rises to a height 'h'. How high is the object in terms of h, when it has lost one third of its original kinetic energy?
Data:
Final Height = h
Height where 1/3 K.E lost (hA) = ?
Solution:
According to the Law of conservation of Energy, the total initial Kinetic Energy at the bottom equals the maximum Potential Energy at the highest point:
K.E. (initial) = m × g × h
Similarly, at any point A, the loss in Kinetic Energy is equal to the gain in Potential Energy:
Loss in K.E. = P.E. at point A
Loss in K.E. = m × g × hA ---(i)
According to the given condition:
Loss in K.E. = 1/3 × K.E. (initial)
So Equation (i) becomes:
1/3 × K.E. (initial) = m × g × hA
1/3 × (m × g × h) = m × g × hA
Canceling (m × g) from both sides:
1/3 × h = hA
hA = h / 3
RESULT: The object will loose its One Third K.E when it has covered one-third of its height.
4. A 70 kg man runs up a hill through a height of 3 m in 2 seconds. (a) How much work does he do against gravitational field? (b) What is the average power output?
Data:
Mass (m) = 70 kg
Height (h) = 3 m
Time (t) = 2 s
Acceleration due to gravity (g) = 9.8 m/s²
Work Done (W) = ?
Power (P) = ?
Solution:
According to the definition of work done against gravitational field:
W = m × g × h
W = 70 × 9.8 × 3
W = 2058 J
According to the definition of Power:
P = W / t
P = 2058 / 2
P = 1029 W
RESULT: Work done by the man is 2058 J against gravitational field and the average power is 1029 W.
5. A neutron travels a distance of 12 m in a time interval of 3.6 x 10⁻⁴ sec. Assuming its speed was constant, Find its kinetic energy? Take the mass of neutron 1.7 x 10⁻²⁷ kg.
Data:
Distance (d) = 12 m
Time (t) = 3.6 × 10⁻⁴ s
Mass of neutron (m) = 1.7 × 10⁻²⁷ kg
Kinetic Energy (K.E) = ?
Solution:
According to the definition of Kinetic Energy:
K.E. = 0.5 × m × v² ---(i)
According to the definition of Velocity:
v = d / t
v = 12 / (3.6 × 10⁻⁴) = 3.33 × 10⁴ m/s
Putting values in eq(i):
K.E. = 0.5 × (1.7 × 10⁻²⁷) × (3.33 × 10⁴)²
K.E. = 9.44 × 10⁻¹⁹ J
For converting into electron volts (1 eV = 1.6 × 10⁻¹⁹ J):
K.E. = (9.44 × 10⁻¹⁹) / (1.6 × 10⁻¹⁹)
K.E. = 5.9 eV
RESULT: The Kinetic energy of neutron is 5.9 eV.
6. A stone is thrown vertically upwards and can reach to a height of 10m, find the speed of stone, when it is just 2m above the ground?
Data:
Maximum Height (hmax) = 10 m
Height above ground (h) = 2 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity at height h (v) = ?
Solution:
According to the Law of Conservation of Energy, total energy at the peak equals total energy at 2 meters:
Total Energy (at top) = Total Energy (at 2m)
m × g × hmax = m × g × h + 0.5 × m × v²
Dividing by mass (m) on both sides:
g × hmax = g × h + 0.5 × v²
Alternatively, looking at the descent via the 3rd Equation of Motion from the maximum height down to 2m (where distance s = 10m - 2m = 8m):
2 × g × s = vf² - vi²
2 × 9.8 × (10 - 2) = v² - 0
19.6 × 8 = v²
v² = 156.8
v = √156.8 = 12.52 m/s
RESULT: The speed of stone 2 m above the ground is 12.5 m/s.
7. The potential energy of a body at the top of a building is 200 Joules when it is dropped its kinetic energy just before striking the ground is 160 Joules, find the work done against the air resistance?
Data:
Potential Energy at top (P.E) = 200 J
Kinetic Energy at bottom (K.E) = 160 J
Work done against air resistance (Wf) = ?
Solution:
According to the definition of Air Resistance and Energy Conservation:
Loss in P.E. = Gain in K.E. + Work done against Air Resistance
200 = 160 + Wf
Wf = 200 - 160
Wf = 40 J
RESULT: The work done against air resistance is 40 J.
8. Find the energy equivalent of 1 gram?
Data:
Mass (m) = 1 g = 1 × 10⁻³ kg
Speed of Light (c) = 3 × 10⁸ m/s
Energy (E) = ?
Solution:
According to the Energy-Mass Relation:
E = m × c²
E = (1 × 10⁻³) × (3 × 10⁸)²
E = 1 × 10⁻³ × 9 × 10¹⁶
E = 9 × 10¹³ J
RESULT: The energy equivalent to 1 gram is 9 × 10¹³ J.
9. A 1 Kilowatt motor pump, pumps the water from the ground to a height of 10 m. Find, how much litres of water it can pump in one hour?
Data:
Power (P) = 1 kW = 1000 W
Height (h) = 10 m
Time (t) = 1 hour = 3600 s
Acceleration due to gravity (g) = 10 m/s²
Volume of water in Litres = ?
Solution:
According to the definition of Power:
P = W / t = (m × g × h) / t
Rearranging the formula to find mass (m):
m = (P × t) / (g × h)
m = (1000 × 3600) / (10 × 10)
m = 3600000 / 100 = 36000 kg
According to the definition of Density, 1 kg of water corresponds to a volume of exactly 1 Litre:
Volume = 36000 Litres
RESULT: The No. of litres that can be pumped is 36000 litres.
10. A rocket of mass 2kg is launched in air, when it attains height of 15m the 400 Joules of its chemical fuel burns. Find the speed of rocket at maximum height?
Data:
Mass of rocket (m) = 2 kg
Height reached (h) = 15 m
Chemical fuel energy converted = 400 J
Acceleration due to gravity (g) = 10 m/s²
Speed (v) = ?
Solution:
The total chemical energy converts into mechanical energy (Potential Energy + Kinetic Energy):
Total Energy = P.E. + K.E.
400 = (m × g × h) + (0.5 × m × v²)
400 = (2 × 10 × 15) + (0.5 × 2 × v²)
400 = 300 + v²
v² = 400 - 300
v² = 100
Taking square root on both sides:
v = √100 = 10 m/s
RESULT: The speed of rocket at maximum height is 10 m/s.
11. A motor pumps the water at the rate 500 gram/minute to the height of 120 m. If the motor is 50% efficient then how much input electric power is needed?
No comments:
Post a Comment