(i) False: Work done requires displacement ($W = F \cdot s$). If $s = 0$, work done is zero.
(ii) True: The upward lifting force and upward displacement are in the same direction.
(iii) True: Both work and energy are measured in Joules ($\text{J}$).
(iv) False: A motionless stretched rubber band has elastic potential energy, not kinetic energy.
(v) True: Energy obeys the Law of Conservation of Energy and can transform between mechanical, thermal, electrical, and other forms.
(i) Work done = Force × Displacement
(ii) 1 joule of work is done when a force of 1 newton displaces an object by 1 metre in the direction of the force.
(iii) The expression for kinetic energy is $\frac{1}{2}mv^2$.
(iv) The potential energy at height h is $mgh$.
(v) Power is defined as the rate at which work is done.
Correct Statements: (iii) Its kinetic energy is zero, and (iv) Its potential energy is maximum.
Reasoning: At the highest point, instantaneous velocity $v = 0 \implies E_k = \frac{1}{2}mv^2 = 0$. Height $h$ is maximum $\implies E_p = mgh$ is maximum. Gravitational force ($F=mg$) and downward acceleration due to gravity ($g=9.8\text{ m/s}^2$) remain non-zero throughout the flight.
Given: $m = 50 \text{ kg}$, $h = 72.5 \text{ m}$, $g = 10 \text{ m/s}^2$
(i) Potential energy gain via elevator:
$$\Delta E_p = mgh = 50 \text{ kg} \times 10 \text{ m/s}^2 \times 72.5 \text{ m} = \mathbf{36,250\text{ J} = 36.25\text{ kJ}}$$(ii) Potential energy gain via staircase:
$$\Delta E_p = mgh = 50 \text{ kg} \times 10 \text{ m/s}^2 \times 72.5 \text{ m} = \mathbf{36,250\text{ J} = 36.25\text{ kJ}}$$(iii) Conclusion:
Gravitational potential energy depends only on the vertical height ($h$) above the reference level and is completely independent of the path taken (conservative force field).
Let height of 1 floor be $h_0 \implies h_{10} = 10 h_0$ and $h_{20} = 20 h_0$. Let time to 10th floor be $t \implies$ time to 20th floor $= 2t$.
1. Energy Comparison:
$$E_{10} = mg(10h_0) = 10 mgh_0$$ $$E_{20} = mg(20h_0) = 20 mgh_0 = 2 \times E_{10}$$The crane requires twice ($2\times$) the energy to lift the mass to the 20th floor.
2. Power Comparison:
$$P_{10} = \frac{E_{10}}{t} = \frac{10 mgh_0}{t}$$ $$P_{20} = \frac{E_{20}}{2t} = \frac{20 mgh_0}{2t} = \frac{10 mgh_0}{t} = P_{10}$$The power requirement remains exactly the same ($1\times$) because double the work is done in double the time.
1. Factors Determining Energy:
Mass of the flag ($m$), height of the flagpole ($h$), and acceleration due to gravity ($g$). Energy required $= mgh$.
2. Speed Effect on Work Done:
No, raising the flag slowly or quickly does not change the work done ($W = mgh$). Work depends only on force and displacement, not on time.
3. Doubling Speed Effect on Power:
Power is $P = \frac{W}{t} = F \cdot v$. If the speed $v$ is doubled, the time required to reach the top is halved ($\frac{t}{2}$). Therefore, the power requirement doubles ($2\times$).
Day 1 (Man alone):
$$\text{Total Mass } m_1 = 60 \text{ kg (man)} + 100 \text{ kg (scooter)} = 160 \text{ kg}$$ $$\text{Energy used } E_1 = \frac{1}{2} m_1 v^2 = \frac{1}{2}(160)v^2 = 80 v^2$$Day 2 (Man + Son):
$$\text{Total Mass } m_2 = 60 \text{ kg} + 40 \text{ kg (son)} + 100 \text{ kg} = 200 \text{ kg}$$ $$\text{Energy used } E_2 = \frac{1}{2} m_2 v^2 = \frac{1}{2}(200)v^2 = 100 v^2$$Ratio of Fuel / Energy Used ($E_1 : E_2$):
$$\frac{\text{Fuel Day 1}}{\text{Fuel Day 2}} = \frac{80 v^2}{100 v^2} = \frac{80}{100} = \mathbf{\frac{4}{5} = 4 : 5}$$Principle of Moments: $\text{Load} \times \text{Load Arm} = \text{Effort} \times \text{Effort Arm}$
Let child mass be $m$, adult mass be $2m$.
$$m \times d_{child} = 2m \times d_{adult} \implies d_{child} = 2 d_{adult}$$To balance the seesaw, the child must sit at twice the distance ($2d$) from the fulcrum compared to the adult sitting at distance $d$.
(i) Sign of Work Done by Gravity:
(ii) Work Done by Air Resistance ($W_{air}$):
Initial Kinetic Energy at launch: $E_{k_i} = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times (20)^2 = 400 \text{ J}$
Potential Energy gained at maximum height ($19.4\text{ m}$): $E_p = mgh = 2 \times 10 \times 19.4 = 388 \text{ J}$
By Work-Energy Theorem: $E_{k_i} + W_{air} = E_p \implies 400 + W_{air} = 388$
$$W_{air} = 388 - 400 = \mathbf{-12\text{ J}}$$Given: $m = 10.0 \text{ kg}$, Initial $E_{k_i} = 180 \text{ J}$ at $s=0\text{ m}$.
(i) Speed at 0 m ($u$):
$$\frac{1}{2} m u^2 = 180 \implies \frac{1}{2}(10)u^2 = 180 \implies 5u^2 = 180 \implies u^2 = 36 \implies u = \mathbf{6\text{ m/s}}$$(ii) Speed at 4 m ($v$):
Work done by variable force = Area under $F-s$ graph (Trapezium from $0\text{ m}$ to $4\text{ m}$, height $F=50\text{ N}$):
$$\text{Area} = \frac{\text{parallel sides sum}}{2} \times \text{height} = \frac{(4 - 0) + (3 - 1)}{2} \times 50 = \frac{4 + 2}{2} \times 50 = 3 \times 50 = 150 \text{ J}$$ $$E_{k_f} = E_{k_i} + W = 180 \text{ J} + 150 \text{ J} = 330 \text{ J}$$ $$\frac{1}{2} (10) v^2 = 330 \implies 5 v^2 = 330 \implies v^2 = 66 \implies v = \sqrt{66} \approx \mathbf{8.12\text{ m/s}}$$Negative Acceleration Query:
No. The force $F$ stays positive ($>0$) throughout $0\text{ m}$ to $4\text{ m}$. Even between $3\text{ m}$ and $4\text{ m}$ where force decreases, $F$ remains positive, so acceleration remains positive ($a > 0$).
Using Kinematic Formula ($v^2 = u^2 - 2gh \implies u^2 = 2gh$ at max height):
Since the initial upward velocity $u$ is the same on Earth and Moon:
$$2 g_{Earth} h_{Earth} = 2 g_{Moon} h_{Moon}$$Given $g_{Moon} = \frac{1}{6} g_{Earth}$ and $h_{Earth} = 8 \text{ m}$:
$$g_{Earth} \times 8 = \left(\frac{1}{6} g_{Earth}\right) \times h_{Moon}$$ $$h_{Moon} = 8 \times 6 = \mathbf{48\text{ m}}$$The ball will reach a height of 48 m on the Moon.
From Fig. 7.38: $m = 1000 \text{ kg}$, speed at A & B $= 35 \text{ m/s}$, reaction time A to B $= 1 \text{ s}$, braking time B to C $= 3 - 1 = 2 \text{ s}$.
(i) Motion between A and B:
The car moves with a constant speed of 35 m/s for 1 second during the driver's reaction time ($a = 0$).
(ii) Kinetic energy at A ($E_k$):
$$E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1000 \times (35)^2 = 500 \times 1225 = \mathbf{612,500\text{ J} = 612.5\text{ kJ}}$$(iii) Work done by brakes (B to C):
$$W = \Delta E_k = 0 - 612,500 \text{ J} = \mathbf{-612,500\text{ J} = -612.5\text{ kJ}}$$(iv) Energy Transformation:
The kinetic energy transforms into thermal (heat) energy in the brake pads and tires, and sound energy (squealing tires).
Given: $m = 0.5 \text{ kg}$, Total Mechanical Energy $E_{total} = E_k(O) + U(O) = 0 + 30 \text{ J} = 30 \text{ J}$.
1. Velocity at P ($U_P = 20\text{ J}$):
$$E_{k_P} = E_{total} - U_P = 30 - 20 = 10 \text{ J}$$ $$\frac{1}{2}(0.5) v_P^2 = 10 \implies 0.25 v_P^2 = 10 \implies v_P^2 = 40 \implies v_P = \sqrt{40} \approx \mathbf{6.32\text{ m/s}}$$2. Velocity at Q ($U_Q = 28\text{ J}$):
$$E_{k_Q} = E_{total} - U_Q = 30 - 28 = 2 \text{ J}$$ $$\frac{1}{2}(0.5) v_Q^2 = 2 \implies 0.25 v_Q^2 = 2 \implies v_Q^2 = 8 \implies v_Q = \sqrt{8} \approx \mathbf{2.83\text{ m/s}}$$3. Velocity at R ($U_R = 40\text{ J}$):
Since $U_R (40\text{ J}) > E_{total} (30\text{ J})$, the ball cannot reach point R. Its velocity reaches 0 m/s and turns back at $U = 30\text{ J}$.
Given: $m = 1.5 \text{ kg}$, $h = 10 \text{ m}$, $F_{resistive} = 3000 \text{ N}$, $g = 10 \text{ m/s}^2$
(i) Velocity just before hitting sand ($v$):
$$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 10} = \sqrt{200} \approx \mathbf{14.14\text{ m/s}}$$(ii) Depth of depression ($d$):
Total energy of coconut at top = $mgh = 1.5 \times 10 \times 10 = 150 \text{ J}$
Work done by resistive force of sand = Total energy $\implies F_{resistive} \times d = 150\text{ J}$
$$3000 \times d = 150 \implies d = \frac{150}{3000} = 0.05 \text{ m} = \mathbf{5\text{ cm}}$$