8. (a) (i) draw a pulley system that has a velocity ratio of 4.
(ii) a body of mass 25 kg is pulled over a rough surface with a 35 n force. if the object accelerated at a rate of 1.5 m⁻², calculate the frictional force acting on the object and the surface. 5 marks
(b) define the following terms as applied to machines:
(i) velocity ratio;
(ii) efficiency. 4 marks
(c) (i) a screw jack of pitch 2 mm is to be used to lift a car of mass 8000 kg. the length of the tommy bar of the jack is 25 cm. calculate the effort that would have been required to attain an efficiency of 85 %.
(ii) explain why the efficiency of a machine is always less than 100 %. 6 marks
9. (a) (i) define the term thermal conductivity.
(ii) give one difference between latent heat of fusion and latent heat of vaporization. 4 marks
(b) (i) an iron rod of mass 2.5 kg at 250 °c is dropped into some quantity of water initially at 33 °c. what would be the mass of the water when the temperature is at 72 °c?
(ii) a piece of ice at −15 °c is subjected to heat until the ice changes to steam at 100 °c. sketch a heating curve to illustrate the changes in temperature during the process. 7 marks
(c) (i) state two effects of heat on a substance.
(ii) by how much should water of temperature −25 °c be increased to obtain its freezing point temperature?

Question

Accepted Answer

### 8 (a)(ii) Solution (Physics - Mechanics)
# Explanation:
## Step1: Recall Newton's Second Law
Newton's second law is $ F_{net} = ma $, where $ F_{net} $ is the net force, $ m $ is mass, and $ a $ is acceleration. The net force is also $ F_{applied}-F_{friction} $, so $ F_{applied}-F_{friction}=ma $.
## Step2: Calculate Net Force
Given $ m = 25\space kg $, $ a = 1.5\space m/s^2 $, so $ F_{net}=25\times1.5 = 37.5\space N $.
## Step3: Solve for Frictional Force
We know $ F_{applied}=35\space N $, and $ F_{applied}-F_{friction}=F_{net} $. Rearranging, $ F_{friction}=F_{applied}-F_{net} $. Wait, but $ F_{applied}=35\space N $ is less than $ F_{net}=37.5\space N $? Wait, no, I made a mistake. Wait, the applied force is 35 N, mass 25 kg, acceleration 1.5 m/s². So net force $ F_{net}=ma = 25\times1.5 = 37.5\space N $. But the applied force is 35 N, which is less than net force? That can't be. Wait, no, the frictional force opposes the motion, so the correct formula is $ F_{applied}-F_{friction}=ma $ only if applied force is in the direction of motion and friction opposite. But if $ F_{applied} < ma $, that would imply friction is negative, which doesn't make sense. Wait, maybe I mixed up. Wait, the net force is $ F_{applied}-F_{friction} $ (if applied force is greater than friction) or $ F_{friction}-F_{applied} $ (if friction is greater). But since the object is accelerating, the net force must be in the direction of motion, so $ F_{applied}-F_{friction}=ma $. But here $ F_{applied}=35\space N $, $ ma = 37.5\space N $, so $ 35 - F_{friction}=37.5 $ would give $ F_{friction}=35 - 37.5=-2.5\space N $, which is impossible. Wait, maybe the acceleration is 1.5 m/s², but maybe I miscalculated. Wait, 25*1.5 is 37.5. So perhaps the applied force is 35 N, but the net force is 37.5 N, which would mean friction is in the opposite direction of applied force? No, that can't be. Wait, maybe the problem has a typo, or I misread. Wait, the mass is 25 kg, force 35 N, acceleration 1.5 m/s². Let's re-express: $ F_{net}=ma = 25\times1.5 = 37.5\space N $. The applied force is 35 N, so the frictional force must be $ F_{friction}=F_{applied}-F_{net}=35 - 37.5=-2.5\space N $. But friction can't be negative. That suggests that maybe the acceleration is in the opposite direction, or the applied force is less than the force needed to overcome friction, but then acceleration would be negative (deceleration). Wait, maybe the problem meant deceleration? Or maybe I made a mistake. Wait, let's check again. Newton's second law: $ \sum F = ma $. If the object is moving, and there's an applied force $ F $ and friction $ f $ (opposing motion), then $ F - f = ma $ (if $ F > f $, acceleration in direction of F) or $ f - F = ma $ (if $ f > F $, acceleration opposite to F, i.e., deceleration). So if the object is accelerating at 1.5 m/s², then $ F - f = ma $ implies $ f = F - ma $. But if $ F = 35\space N $, $ ma = 37.5\space N $, then $ f = 35 - 37.5=-2.5\space N $, which is impossible. So maybe the acceleration is 1.0 m/s²? Or the mass is 20 kg? Wait, the problem says 25 kg, 35 N, 1.5 m/s². Maybe the question is correct, and the negative sign indicates that the friction is in the same direction as the applied force? No, friction opposes relative motion. So perhaps there's a mistake in the problem, but assuming the problem is correct, maybe we proceed. Wait, maybe I messed up the formula. Let's do it again: $ F_{net}=ma = 25\times1.5 = 37.5\space N $. The net force is the vector sum of applied force and friction. So $ \vec{F}_{net}=\vec{F}_{applied}+\vec{F}_{friction} $. If the applied force is in the positive direction, and friction is in the negative direction, then $ F_{net}=F_{applied}-F_{friction} $. So $ 37.5 = 35 - F_{friction} $ → $ F_{friction}=35 - 37.5=-2.5\space N $. The negative sign means that the frictional force is actually in the positive direction, i.e., it's aiding the motion, which is possible if there's another force, but the problem says "pulled over a rough surface with a 35 N force", so maybe the surface is moving, or there's a mistake. Alternatively, maybe the acceleration is 1.0 m/s². Wait, maybe the user made a typo, but assuming the problem is as stated, the frictional force is -2.5 N, which is equivalent to 2.5 N in the direction of motion. But that's unusual. Alternatively, maybe I made a mistake in the sign. Let's consider that the net force is $ F_{friction}-F_{applied}=ma $ (if friction is greater, causing deceleration, but the problem says "accelerated", so acceleration is in the direction of applied force). So this is confusing. Maybe the problem has a mistake, but proceeding with the given numbers: $ F_{friction}=F_{applied}-ma = 35 - (25\times1.5)=35 - 37.5=-2.5\space N $. So the frictional force is 2.5 N in the direction of the applied force (which is unusual for friction, but mathematically that's the result).

# Answer:
The frictional force is $ \boldsymbol{-2.5\space N} $ (or 2.5 N in the direction of the applied force, indicating a possible error in problem parameters).

### 8 (b) Solutions
#### (i) Velocity Ratio (VR)
# Brief Explanations:
Velocity ratio of a machine is the ratio of the velocity of the effort ($ v_E $) to the velocity of the load ($ v_L $) or the ratio of the distance moved by the effort ($ d_E $) to the distance moved by the load ($ d_L $) in the same time. Mathematically, $ VR=\frac{v_E}{v_L}=\frac{d_E}{d_L} $.
# Answer:
Velocity ratio (VR) of a machine is the ratio of the velocity of the effort to the velocity of the load (or the ratio of the distance moved by the effort to the distance moved by the load) in the same time interval.

#### (ii) Efficiency ($ \eta $)
# Brief Explanations:
Efficiency of a machine is the ratio of the useful work output ($ W_{out} $) to the total work input ($ W_{in} $) expressed as a percentage. It can also be defined as the ratio of mechanical advantage (MA) to velocity ratio (VR) times 100, i.e., $ \eta=\frac{W_{out}}{W_{in}}\times100\%=\frac{MA}{VR}\times100\% $.
# Answer:
Efficiency of a machine is the percentage ratio of the useful work output to the total work input (or the ratio of mechanical advantage to velocity ratio, multiplied by 100%).

### 8 (c)(i) Solution (Physics - Machines)
# Explanation:
## Step1: Recall Formulas for Screw Jack
For a screw jack, the velocity ratio $ VR=\frac{2\pi L}{p} $, where $ L $ is the length of the tommy bar, and $ p $ is the pitch. Efficiency $ \eta=\frac{MA}{VR}\times100\% $, and mechanical advantage $ MA=\frac{Load}{Effort}=\frac{mg}{E} $, where $ m $ is mass, $ g = 9.8\space m/s^2 $, $ E $ is effort.
## Step2: Calculate Velocity Ratio
Given $ L = 25\space cm = 0.25\space m $, $ p = 2\space mm = 0.002\space m $. So $ VR=\frac{2\pi\times0.25}{0.002}=\frac{0.5\pi}{0.002}=250\pi\approx785.4 $.
## Step3: Calculate Mechanical Advantage from Efficiency
Efficiency $ \eta = 85\% = 0.85 $. From $ \eta=\frac{MA}{VR}\times100\% $, we get $ MA=\frac{\eta\times VR}{100\%} $. Wait, no: $ \eta=\frac{MA}{VR}\times100\% $ → $ MA=\frac{\eta\times VR}{100\%} $? No, $ \eta=\frac{MA}{VR}\times100 $ (in percentage), so $ MA=\frac{\eta\times VR}{100} $. Wait, no, $ \eta=\frac{W_{out}}{W_{in}}=\frac{Load\times d_L}{Effort\times d_E}=\frac{MA}{VR} $ (since $ MA=\frac{Load}{Effort} $, $ VR=\frac{d_E}{d_L} $), so $ \eta=\frac{MA}{VR} $ (as a decimal), so $ MA=\eta\times VR $.
## Step4: Calculate Load
Load $ = mg = 8000\space kg\times9.8\space m/s^2 = 78400\space N $.
## Step5: Solve for Effort
From $ MA=\frac{Load}{Effort} $, we have $ Effort=\frac{Load}{MA} $. And $ MA=\eta\times VR $, so $ Effort=\frac{Load}{\eta\times VR} $. Substituting values: $ VR = \frac{2\pi L}{p}=\frac{2\pi\times0.25}{0.002}=\frac{0.5\pi}{0.002}=250\pi\approx785.4 $. $ \eta = 0.85 $, $ Load = 78400\space N $. So $ Effort=\frac{78400}{0.85\times785.4}\approx\frac{78400}{667.59}\approx117.4\space N $.
# Answer:
The effort required is approximately $ \boldsymbol{117.4\space N} $ (or more precisely, using $ \pi\approx3.1416 $, $ VR = 250\times3.1416 = 785.4 $, $ Effort=\frac{8000\times9.8}{0.85\times785.4}\approx\frac{78400}{667.59}\approx117.4\space N $).

### 8 (c)(ii) Explanation
# Brief Explanations:
The efficiency of a machine is always less than 100% because of energy losses. These losses occur due to friction between moving parts (converting energy to heat), air resistance, deformation of machine parts, and sound energy. Some of the input work is used to overcome these resistive forces (non - useful work), so the useful work output is always less than the work input.
# Answer:
The efficiency of a machine is always less than 100% because some of the input work is used to overcome friction (producing heat), air resistance, and other energy - dissipating factors (like deformation of parts). Thus, the useful work output is less than the total work input.

### 9 (a)(i) Definition of Thermal Conductivity
# Brief Explanations:
Thermal conductivity of a material is the rate of heat transfer (heat energy per unit time) through a unit cross - sectional area of the material, per unit temperature gradient (per unit change in temperature per unit distance) across the material. Mathematically, $ k=\frac{Qd}{At\Delta T} $, where $ k $ is thermal conductivity, $ Q $ is heat transferred, $ d $ is thickness, $ A $ is area, $ t $ is time, and $ \Delta T $ is temperature difference.
# Answer:
Thermal conductivity of a material is the rate of heat transfer through a unit cross - sectional area of the material, per unit temperature gradient (per unit temperature change per unit distance) across it.

### 9 (a)(ii) Difference between Latent Heat of Fusion and Vaporization
# Brief Explanations:
- Latent heat of fusion ($ L_f $) is the heat energy required to change a unit mass of a substance from the solid state to the liquid state at its melting point (without a change in temperature). For example, $ L_f $ of water is the heat needed to melt ice at $ 0^{\circ}C $ to water at $ 0^{\circ}C $.
- Latent heat of vaporization ($ L_v $) is the heat energy required to change a unit mass of a substance from the liquid state to the gaseous state at its boiling point (without a change in temperature). For example, $ L_v $ of water is the heat needed to boil water at $ 100^{\circ}C $ to steam at $ 100^{\circ}C $.
- A key difference is that $ L_v $ is generally much larger than $ L_f $ for most substances (e.g., for water, $ L_v\approx2.26\times10^6\space J/kg $ and $ L_f\approx3.34\times10^5\space J/kg $).
# Answer:
- Latent heat of fusion is the heat to change 1 kg of a solid to liquid at its melting point (no temp change). Latent heat of vaporization is the heat to change 1 kg of a liquid to gas at its boiling point (no temp change).
- $ L_v $ is usually much larger than $ L_f $ (e.g., for water, $ L_v\approx2.26\times10^6\space J/kg $, $ L_f\approx3.34\times10^5\space J/kg $).

### 9 (b)(i) Solution (Heat Transfer - Calorimetry)
# Explanation:
## Step1: Recall Principle of Calorimetry
Heat lost by the iron rod = Heat gained by the water. $ Q_{lost}=m_{iron}c_{iron}(T_{iron,initial}-T_{final}) $, $ Q_{gained}=m_{water}c_{water}(T_{final}-T_{water,initial}) $. Assume $ c_{iron}=450\space J/(kg\cdot^{\circ}C) $, $ c_{water}=4200\space J/(kg\cdot^{\circ}C) $.
## Step2: Calculate Heat Lost by Iron
$ m_{iron}=2.5\space kg $, $ T_{iron,initial}=250^{\circ}C $, $ T_{final}=72^{\circ}C $, $ c_{iron}=450\space J/(kg\cdot^{\circ}C) $. So $ Q_{lost}=2.5\times450\times(250 - 72)=2.5\times450\times178 = 2.5\times80100 = 200250\space J $.
## Step3: Solve for Mass of Water
Let $ m_{water}=m $. $ Q_{gained}=m\times4200\times(72 - 33)=m\times4200\times39 = m\times163800 $. Since $ Q_{lost}=Q_{gained} $, $ 200250 = 163800m $. So $ m=\frac{200250}{163800}\approx1.22\space kg $.
# Answer:
The mass of the water is approximately $ \boldsymbol{1.22\space kg} $ (using $ c_{iron}=450\space J/(kg\cdot^{\circ}C) $ and $ c_{water}=4200\space J/(kg\cdot^{\circ}C) $).

### 9 (b)(ii) Heating Curve for Ice to Steam
# Brief Explanations:
- **Stage 1: Heating ice from $- 15^{\circ}C$ to $0^{\circ}C$ (Solid heating):** Temperature increases linearly with time (positive slope) as heat is added. The x - axis is time, y - axis is temperature.
- **Stage 2: Melting of ice at $0^{\circ}C$ (Phase change - solid to liquid):** Temperature

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

8 (a)(ii) Solution (Physics - Mechanics)

Step1: Recall Newton's Second Law

Step2: Calculate Net Force

Step3: Solve for Frictional Force

Answer:

8 (b) Solutions

(i) Velocity Ratio (VR)