Work, Energy, and Simple MachinesClass 9 Science NCERT Solutions

Let the height of one floor be $h$. Let the mass be $m$ and acceleration due to gravity be $g$.

Case 1: Lifting to the 10th floor Height, $H_1 = 10h$ Let the time taken be $t_1 = t$.

Energy required (Work done), $E_1 = m g H_1 = mg(10h) = 10mgh$ Power required, $P_1 = \frac{E_1}{t_1} = \frac{10mgh}{t}$

Case 2: Lifting to the 20th floor Height, $H_2 = 20h$ Time taken, $t_2 = 2t$ (double the time).

Energy required (Work done), $E_2 = m g H_2 = mg(20h) = 20mgh$ Power required, $P_2 = \frac{E_2}{t_2} = \frac{20mgh}{2t} = \frac{10mgh}{t}$

Final Answer: To lift the mass to the 20th floor, twice the energy is required, but the power required is the same.

Factors determining the energy required: The energy required to raise the flag is equal to the work done against gravity, which is stored as gravitational potential energy. The factors that determine this energy are:

Effect of speed on work done: Raising the flag slowly or quickly does not change the amount of work done against gravity. The work done is determined by the force (weight of the flag, $mg$) and the vertical displacement (height of the pole, $h$). As long as the initial and final heights are the same, the work done against gravity remains $mgh$, regardless of the time taken.

Effect of speed on power requirement: Power is the rate at which work is done ($P = \frac{W}{t}$). If the speed at which the flag is raised is doubled, the time ($t$) taken to cover the same height ($h$) will be halved. Let the initial power be $P_1 = \frac{W}{t}$. When speed is doubled, the new time is $t_2 = \frac{t}{2}$. The new power requirement will be $P_2 = \frac{W}{t_2} = \frac{W}{t/2} = 2 \times (\frac{W}{t}) = 2P_1$. Therefore, if the speed is doubled, the power requirement is also doubled because the same amount of work must be done in half the time.

The energy from the fuel is converted into the kinetic energy of the scooter and its riders. We need to find the ratio of the kinetic energies on the two days, as this will be the ratio of the fuel used.

Day 1: Man rides alone Given: Mass of the man, $m_{man} = 60 \text{ kg}$ Mass of the scooter, $m_{scooter} = 100 \text{ kg}$ Total mass on Day 1, $M_1 = m_{man} + m_{scooter} = 60 \text{ kg} + 100 \text{ kg} = 160 \text{ kg}$ Final velocity, $v$

Calculation of Energy on Day 1: Kinetic energy, $K_1 = \frac{1}{2} M_1 v^2 = \frac{1}{2} \times 160 \times v^2 = 80v^2$ Fuel used on Day 1 is proportional to $K_1$.

Day 2: Man rides with his son Given: Mass of the son, $m_{son} = 40 \text{ kg}$ Total mass on Day 2, $M_2 = m_{man} + m_{son} + m_{scooter} = 60 \text{ kg} + 40 \text{ kg} + 100 \text{ kg} = 200 \text{ kg}$ Final velocity, $v$

Calculation of Energy on Day 2: Kinetic energy, $K_2 = \frac{1}{2} M_2 v^2 = \frac{1}{2} \times 200 \times v^2 = 100v^2$ Fuel used on Day 2 is proportional to $K_2$.

Ratio of Fuel Used: The ratio of the fuel used on the two days is the ratio of the kinetic energies. Ratio = $\frac{\text{Fuel used on Day 1}}{\text{Fuel used on Day 2}} = \frac{K_1}{K_2} = \frac{80v^2}{100v^2} = \frac{80}{100} = \frac{4}{5}$.

Final Answer: The ratio of the fuel used on the first day to the second day is 4:5.

For the seesaw to be balanced, the principle of moments must apply. The turning effect (moment) on one side of the fulcrum must be equal to the turning effect on the other side.

Principle of Moments: $\text{Effort} \times \text{Effort arm} = \text{Load} \times \text{Load arm}$

Let: Weight of the child = $W_c$ Weight of the adult = $W_a$ Distance of the child from the fulcrum = $L_c$ Distance of the adult from the fulcrum = $L_a$

Given: The adult weighs twice that of the child, so $W_a = 2W_c$.

For the seesaw to be balanced: $W_c \times L_c = W_a \times L_a$ Substituting $W_a = 2W_c$: $W_c \times L_c = (2W_c) \times L_a$ $L_c = 2L_a$

This means the child must sit at a distance from the fulcrum that is twice the distance of the adult from the fulcrum.

Figure Description: A figure would show a horizontal plank (the seesaw) balanced on a triangular pivot (the fulcrum) at its center.

Given: Mass of the ball, $m = 2 \text{ kg}$ Initial velocity, $u = 20 \text{ m s}^{-1}$ Maximum height reached, $h = 19.4 \text{ m}$ Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

(ii) Work done by air resistance: We can use the work-energy theorem, which states that the total work done on an object is equal to the change in its kinetic energy. $W_{total} = \Delta K = K_f - K_i$

The forces doing work on the ball during its upward motion are gravity and air resistance. The final velocity ($v_f$) at the maximum height is 0.

Final Answer: The work done by air resistance was -12 J.

Given: Mass of the block, $m = 10.0 \text{ kg}$ Initial kinetic energy at $x=0$, $K_i = 180 \text{ J}$ The force-displacement graph (Fig. 7.37) shows a force that increases linearly from 0 N at $x=0$ m to 20 N at $x=2$ m, and then decreases linearly back to 0 N at $x=4$ m.

(i) Block's speed at 0 m: Formula: $K = \frac{1}{2}mv^2$ Calculation: $180 \text{ J} = \frac{1}{2} \times 10.0 \text{ kg} \times v_i^2$ $180 = 5 \times v_i^2$ $v_i^2 = \frac{180}{5} = 36$ $v_i = \sqrt{36} = 6 \text{ m s}^{-1}$ Final Answer: The block's speed at 0 m is $6 \text{ m s}^{-1}$.

(ii) Block's speed at 4 m: According to the work-energy theorem, the work done by the force equals the change in kinetic energy. $W = K_f - K_i$ Work done ($W$) is the area under the force-displacement graph.

Calculation of Work Done: The graph is a triangle with base = 4 m and height = 20 N. $W = \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height}$ $W = \frac{1}{2} \times 4 \text{ m} \times 20 \text{ N} = 40 \text{ J}$

Calculation of Final Speed: $W = K_f - K_i$ $40 \text{ J} = K_f - 180 \text{ J}$ $K_f = 40 \text{ J} + 180 \text{ J} = 220 \text{ J}$ Now, find the final speed $v_f$: $K_f = \frac{1}{2}mv_f^2$ $220 \text{ J} = \frac{1}{2} \times 10.0 \text{ kg} \times v_f^2$ $220 = 5 \times v_f^2$ $v_f^2 = \frac{220}{5} = 44$ $v_f = \sqrt{44} \approx 6.63 \text{ m s}^{-1}$ Final Answer: The block's speed at 4 m is approximately $6.63 \text{ m s}^{-1}$.

Negative Acceleration: Acceleration is given by Newton's second law, $a = \frac{F}{m}$. For acceleration to be negative, the net force $F$ must be negative. Looking at the graph (Fig. 7.37), the applied force is always positive (or zero) in the interval from 0 m to 4 m. Since the force is never negative, the acceleration is also never negative.

Final Answer: No, the block does not have negative acceleration in any portion of this motion.

Given: Gravitational acceleration on Moon, $g_m = \frac{1}{6} g_e$ (where $g_e$ is on Earth) Maximum height on Earth, $h_e = 8 \text{ m}$ The ball is thrown with the same initial velocity ($u$) on both Earth and the Moon.

To Find: Maximum height on the Moon, $h_m$.

Method: When the ball is thrown upwards, its initial kinetic energy is converted into gravitational potential energy at its maximum height. By the principle of conservation of energy: Initial Kinetic Energy = Final Potential Energy $\frac{1}{2} m u^2 = mgh$

Since the initial velocity ($u$) is the same in both cases, the initial kinetic energy ($\frac{1}{2} m u^2$) is also the same.

On Earth: $\frac{1}{2} m u^2 = m g_e h_e$

On the Moon: $\frac{1}{2} m u^2 = m g_m h_m$

Calculation: Since both expressions are equal to the same initial kinetic energy, we can equate them: $m g_e h_e = m g_m h_m$ Canceling $m$ from both sides: $g_e h_e = g_m h_m$ Substitute the given values: $g_e \times 8 \text{ m} = (\frac{1}{6} g_e) \times h_m$ Cancel $g_e$ from both sides: $8 \text{ m} = \frac{1}{6} h_m$ $h_m = 8 \text{ m} \times 6 = 48 \text{ m}$

Final Answer: The ball will travel up to a height of 48 m on the surface of the Moon.

Given: Mass of the car, $m = 1000 \text{ kg}$ From the graph (Fig. 7.38): Speed at A and B, $v = 15 \text{ m s}^{-1}$ Speed at C, $v_f = 0 \text{ m s}^{-1}$

(i) Motion between A and B: The graph shows that from time $t=0$ (position A) to $t=1$ s (position B), the speed of the car remains constant at $15 \text{ m s}^{-1}$. This interval represents the driver's reaction time, during which the car continues to move at a constant velocity before the brakes are applied.

(ii) Kinetic energy of the car at A: Formula: $K = \frac{1}{2}mv^2$ Calculation: $K_A = \frac{1}{2} \times 1000 \text{ kg} \times (15 \text{ m s}^{-1})^2$ $K_A = 500 \times 225 = 112500 \text{ J}$ Final Answer: The kinetic energy of the car at A is $112,500$ J.

(iii) Work done by the brakes between B and C: According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. $W_{brakes} = \Delta K = K_C - K_B$ Kinetic energy at B, $K_B = 112500 \text{ J}$ (since speed is the same as at A). Kinetic energy at C, $K_C = 0$ (since the car is at rest). Calculation: $W_{brakes} = 0 - 112500 \text{ J} = -112500 \text{ J}$ Final Answer: The work done by the brakes is -112,500 J. The negative sign indicates that the force applied by the brakes is opposite to the direction of motion.

(iv) Transformation of kinetic energy: When the brakes are applied, the kinetic energy of the car is converted primarily into heat energy due to the friction between the brake pads and the wheels, and between the tires and the road surface. A small amount of energy is also converted into sound energy (the screeching of tires).

Given: Mass of the ball, $m = 0.5 \text{ kg}$ At position O, velocity $v_O = 0 \text{ m s}^{-1}$ and potential energy $U_O = 30 \text{ J}$. The track is frictionless, so total mechanical energy is conserved.

Total Mechanical Energy: The total mechanical energy ($E_{total}$) of the system is the sum of its kinetic and potential energy. It remains constant. $E_{total} = K_O + U_O = 0 + 30 \text{ J} = 30 \text{ J}$

Velocity at P: From the graph (Fig. 7.39), the potential energy at P is $U_P = 10 \text{ J}$. Using conservation of energy: $E_{total} = K_P + U_P$ $30 \text{ J} = K_P + 10 \text{ J}$ $K_P = 20 \text{ J}$ Now, find the velocity $v_P$: $K_P = \frac{1}{2}mv_P^2 \Rightarrow 20 = \frac{1}{2} \times 0.5 \times v_P^2$ $20 = 0.25 \times v_P^2 \Rightarrow v_P^2 = \frac{20}{0.25} = 80$ $v_P = \sqrt{80} \approx 8.94 \text{ m s}^{-1}$

Velocity at Q: From the graph, the potential energy at Q is $U_Q = 20 \text{ J}$. Using conservation of energy: $E_{total} = K_Q + U_Q$ $30 \text{ J} = K_Q + 20 \text{ J}$ $K_Q = 10 \text{ J}$ Now, find the velocity $v_Q$: $K_Q = \frac{1}{2}mv_Q^2 \Rightarrow 10 = \frac{1}{2} \times 0.5 \times v_Q^2$ $10 = 0.25 \times v_Q^2 \Rightarrow v_Q^2 = \frac{10}{0.25} = 40$ $v_Q = \sqrt{40} \approx 6.32 \text{ m s}^{-1}$

Velocity at R: From the graph, the potential energy at R is $U_R = 0 \text{ J}$. Using conservation of energy: $E_{total} = K_R + U_R$ $30 \text{ J} = K_R + 0 \text{ J}$ $K_R = 30 \text{ J}$ Now, find the velocity $v_R$: $K_R = \frac{1}{2}mv_R^2 \Rightarrow 30 = \frac{1}{2} \times 0.5 \times v_R^2$ $30 = 0.25 \times v_R^2 \Rightarrow v_R^2 = \frac{30}{0.25} = 120$ $v_R = \sqrt{120} \approx 10.95 \text{ m s}^{-1}$

Final Answer: The velocity of the ball at P is approximately $8.94 \text{ m s}^{-1}$. The velocity of the ball at Q is approximately $6.32 \text{ m s}^{-1}$. The velocity of the ball at R is approximately $10.95 \text{ m s}^{-1}$.

Given: Mass of the coconut, $m = 1.5 \text{ kg}$ Height of the tree, $h = 10 \text{ m}$ Average resistive force of sand, $F_{sand} = 3000 \text{ N}$ Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

(i) Velocity just before hitting the sand: We use the principle of conservation of energy for the falling coconut. The potential energy at the top of the tree is converted into kinetic energy just before it hits the sand. Potential Energy at top = Kinetic Energy at bottom $mgh = \frac{1}{2}mv^2$ $gh = \frac{1}{2}v^2$ $v^2 = 2gh$ Calculation: $v^2 = 2 \times 10 \text{ m s}^{-2} \times 10 \text{ m} = 200 \text{ m}^2\text{s}^{-2}$ $v = \sqrt{200} = 10\sqrt{2} \approx 14.14 \text{ m s}^{-1}$ Final Answer: The velocity of the coconut just before it hits the sand is approximately $14.14 \text{ m s}^{-1}$.

(ii) Depth of the depression in the sand: Let the depth of the depression be $d$. The total energy of the coconut at the top of the tree (with respect to its final resting position inside the sand) is converted into work done by the sand to stop it.

Total initial energy of the coconut = Potential energy at height $(h+d)$. $E_{total} = mg(h+d)$ Work done by the resistive force of the sand = $F_{sand} \times d$.

According to the work-energy theorem, the work done by all forces equals the change in kinetic energy (which is zero, as it starts and ends at rest). A simpler way is to equate the total energy lost by the coconut to the work done by the sand. $mg(h+d) = F_{sand} \times d$ $mgh + mgd = F_{sand} \times d$ $mgh = F_{sand} \times d - mgd$ $mgh = (F_{sand} - mg)d$ Calculation: $1.5 \text{ kg} \times 10 \text{ m s}^{-2} \times 10 \text{ m} = (3000 \text{ N} - 1.5 \text{ kg} \times 10 \text{ m s}^{-2}) \times d$ $150 \text{ J} = (3000 \text{ N} - 15 \text{ N}) \times d$ $150 = 2985 \times d$ $d = \frac{150}{2985} \approx 0.05025 \text{ m}$ Converting to centimeters: $0.05025 \text{ m} \times 100 = 5.025 \text{ cm}$.

Final Answer: The depth of the depression the coconut makes in the sand is approximately 5.03 cm.