QUESTION 3
Question 3.1 [2 marks] Define the term kinetic frictional force. The kinetic frictional force is the force that opposes the relative motion of two surfaces in contact that are sliding against each other. It acts parallel to the surfaces and is typically proportional to the normal force exerted between the surfaces.
Question 3.2 [2 marks] State Newton's First Law of Motion in words. Newton's First Law of Motion states that an object at rest will remain at rest, and an object in motion will continue to move in a straight line at a constant velocity unless acted upon by a net external force.
Question 3.3 [2 marks] If the person in Figure 2 is applying a force of 275N on the handle, calculate the magnitude of the horizontal component of the applied force. The horizontal component (\(F_{x}\)) of the applied force can be calculated using trigonometry:
\[ F_{x} = F \times \cos(\theta) \] Where:
- \(F = 275N\)
- \(\theta = 50^\circ\)
Calculating:
\[ F_{x} = 275 \times \cos(50^\circ) \approx 275 \times 0.6428 \approx 176.8N \]
Thus, the horizontal component of the applied force is approximately 176.8 N.
Question 3.4 [4 marks] Calculate the magnitude of the normal force. For the reclaimer moving at constant velocity, the net force in the vertical direction is zero. The vertical forces include the weight of the cart \( (mg) \), the vertical component of the applied force, and the normal force (\(N\)).
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Calculate the weight of the cart:
\[ m = 75 kg \quad (mass\ of\ the\ cart) \] \[ g = 9.8 m/s^2 \quad (acceleration\ due\ to\ gravity) \] \[ Weight\ (W) = mg = 75 \times 9.8 = 735 N \]
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Calculate the vertical component of the applied force:
\[ F_{y} = F \times \sin(\theta) = 275 \times \sin(50°) = 275 \times 0.7660 \approx 210.15 N \]
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Set up the equilibrium equation in the vertical direction:
\[ N + F_{y} - W = 0 \] \[ N + 210.15 - 735 = 0 \] \[ N = 735 - 210.15 \approx 524.85 N \]
Thus, the magnitude of the normal force is approximately 524.85 N.
Question 3.5 [3 marks] Calculate the coefficient of kinetic friction. The force of kinetic friction (\(f_k\)) can be expressed as:
\[ f_k = \mu_k N \]
Given that the cart is moving at constant velocity, the net horizontal forces are zero:
\[ F_{x} - f_k = 0 \ f_k = F_{x} \approx 176.8 N \]
Substituting into the kinetic friction formula:
\[ \mu_k = \frac{f_k}{N} = \frac{176.8}{524.85} \approx 0.336 \]
Thus, the coefficient of kinetic friction is approximately 0.336.
Question 3.6 [2 marks] Will the coefficient of kinetic friction change if the angle of the applied force is decreased? Write only YES or NO and explain your answer. NO. The coefficient of kinetic friction depends on the nature of the surfaces in contact, not the angle of the applied force. However, decreasing the angle may affect the normal force and thus change the frictional force, but the coefficient itself remains constant.
Question 3.7 [2 marks] How will the magnitude of kinetic friction change if the angle of the applied force is decreased? Write INCREASE, DECREASE, OR STAY THE SAME. Explain your answer. DECREASE. As the angle of the applied force is decreased, the vertical component of the applied force \(F_y\) decreases, resulting in an increase in normal force \(N\). The kinetic friction force depends on the normal force, so if the normal force increases, the kinetic friction force will generally increase, but if the angle decrease leads to a decrease in the horizontal component \(F_{x}\), the overall kinetic friction force may reduce.
QUESTION 4
Question 4.1.1 [2 marks] Define Newton’s second law of motion. Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is often expressed with the formula \(F = ma\), where \(F\) is the net force, \(m\) is mass, and \(a\) is acceleration.
Question 4.1.2 [4 marks] Draw a labelled free-body diagram for the bag of sugar.
Please imagine a simple free-body diagram with:
- An arrow pointing down labelled as "Weight of sugar (W_s = mg)"
- An arrow pointing up labelled as "Normal force (N)"
- An arrow pointing left labelled as "Friction force (f = 0.01 N)"
- An arrow pointing to the right (a diagonal arrow at angle θ) labelled as "Applied force (F = 0.25 N)"
Question 4.1.3 [6 marks] Calculate the magnitude of Angle θ. First, resolve the forces acting on the system. The net force \( F_{net} \) acting on the mass is defined as:
\[ F_{net} = (m_{sugar} + m_{flour}) \cdot a \]
The total mass (\(m_{sugar} = 750 g = 0.75 kg\), \(m_{flour} = 2.5 kg\)):
\[ F_{net} = (0.75 + 2.5) \cdot 0.05 = 3.25 \cdot 0.05 = 0.1625 N \]
Now, since the net force is the resultant of the applied force, and we know the frictional forces, we can express:
\[ F_{net} = F - (f_{sugar} + f_{flour}) \] Where \( f_{sugar} = 0.01 N \) and \( f_{flour} = 0.02 N \):
Calculating total frictional force:
\[ f_{total} = 0.01 + 0.02 = 0.03 N \]
Then,
\[ 0.1625 = 0.25 \cos(θ) - 0.03 \] \[ 0.1625 + 0.03 = 0.25 \cos(θ) \] \[ 0.1925 = 0.25 \cos(θ) \] \[ \cos(θ) = \frac{0.1925}{0.25} = 0.77 \] \[ θ = \cos^{-1}(0.77) \approx 40^\circ \] Therefore, the magnitude of angle \(θ\) is approximately 40 degrees.
Question 4.2.1 [6 marks] Use your knowledge of Newton’s II calculations to solve for the acceleration of the objects. For the pulley system, assume the two bags are connected and will have the same magnitude of acceleration.
The net force acting on the two bags can be calculated:
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Calculate the effective mass (combined mass): \[ m = m_{sugar} + m_{flour} = 0.75 + 2.5 = 3.25 kg \]
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The tension (T) in the string needs to overcome the friction of both bags also.
Taking downward forces from gravity minus total friction acting against the motion: \[ F_{net} = m \cdot g - (f_{sugar} + f_{flour}) \] Where \(g \approx 9.8 m/s2\):
\[ F_{gravity} = m \cdot g = 3.25 \cdot 9.8 = 31.85 N \] \[ F_{net} = 31.85 - (0.01 + 0.02) = 31.85 - 0.03 = 31.82 N \] Thus,
Now using Newton's Second Law:
\[ F_{net} = ma \] Rearranging,
\[ a = \frac{F_{net}}{m} = \frac{31.82}{3.25} \approx 9.8 m/s^2. \]
Hence, the acceleration of the objects is approximately 9.8 m/s².
Note: Check calculations and measures for any specific figure references if necessary. Adjust units if required.