Past Exam Questions

Resolve forces parallel to the plane:

\(65 \cos 36^{\circ} = 12g \sin 24^{\circ} + F\)

\(F = 65 \cos 36^{\circ} - 12g \sin 24^{\circ} = 3.777707 \ldots\)

Resolve forces perpendicular to the plane:

\(12g \cos 24^{\circ} = R + 65 \sin 36^{\circ}\)

\(R = 12g \cos 24^{\circ} - 65 \sin 36^{\circ} = 71.419 \ldots\)

Use \(F = \mu R\):

\(\mu = \frac{F}{R} = \frac{3.777707}{71.419} = 0.0529\)

(i) Resolve forces parallel to the plane. For the block on the point of sliding down:

\(2X + F = 11g \sin 30^{\circ}\)

For the block on the point of sliding up:

\(9X - F = 11g \sin 30^{\circ}\)

Adding these equations gives:

\(11X = 22g \sin 30^{\circ}\)

\(X = 10\)

(ii) The frictional force \(F\) when the block is on the point of sliding is 35 N. The normal reaction \(R\) is given by:

\(R = 11g \cos 30^{\circ}\)

The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = \frac{35}{11g \cos 30^{\circ}} = 0.367\)

(i) Resolve forces parallel to the plane:

\(F + T = 8 \times 10 \sin 20^{\circ}\)

\(F + 13 = 8 \times 10 \times 0.342\)

\(F = 27.36 - 13 = 14.36 \text{ N}\)

Frictional component is 14.4 N (rounded).

Resolve forces normal to the plane:

\(R = 8 \times 10 \cos 20^{\circ}\)

\(R = 80 \times 0.94 = 75.2 \text{ N}\)

Normal component is 75.2 N.

(ii) When the string is cut, the block is on the point of slipping, so frictional force \(F = 8 \times 10 \sin 20^{\circ}\).

Coefficient of friction \(\mu = \tan 20^{\circ}\).

\(\mu = 0.364\)

First, resolve the forces acting on the block. The gravitational force component parallel to the plane is given by:

\(F_g = mg \sin \theta = 20 \times 9.8 \times \sin 10^\circ\)

\(F_g = 34.2 \text{ N}\)

The normal force \(R\) is given by:

\(R = mg \cos \theta = 20 \times 9.8 \times \cos 10^\circ\)

\(R = 193.2 \text{ N}\)

The frictional force \(F_f\) is:

\(F_f = \mu R = 0.32 \times 193.2 = 61.8 \text{ N}\)

(i) When the force acts up the plane, the equation of motion is:

\(P = F_f + F_g = 61.8 + 34.2 = 96 \text{ N}\)

Thus, the least magnitude of the force is 97.8 N.

(ii) When the force acts down the plane, the equation of motion is:

\(P = F_f - F_g = 61.8 - 34.2 = 27.6 \text{ N}\)

Thus, the least magnitude of the force is 28.3 N.

(i) The normal force \(R\) is given by the component of the weight perpendicular to the plane:

\(R = 15 \times 10 \times \cos 35^{\circ} = 123\)

(ii) For resolving forces along the plane:

When the box is about to move down, the equation is:

\(150 \sin 35^{\circ} = X + F\)

When the box is about to move up, the equation is:

\(150 \sin 35^{\circ} = 5X - F\)

Eliminating \(F\) from these equations gives:

\(X = 28.1\)

For the coefficient of friction \(\mu\), use:

\(F = \mu R\)

\(57.36 = \mu \times 122.9\)

\(\mu = 0.467\)

(a) The forces acting on the particle include:

• The gravitational force \(12g\) acting vertically downward.
• The normal reaction \(R\) perpendicular to the plane.
• The frictional force \(F\) acting up the plane.
• The force \(P\) acting parallel to the plane.

(b) To find the least possible value of \(P\), resolve the forces parallel and perpendicular to the plane:

Parallel to the plane:

\(P + F = 12g \sin 25^{\circ}\)

Perpendicular to the plane:

\(R = 12g \cos 25^{\circ}\)

The frictional force \(F\) is given by:

\(F = \mu R = 0.35 \times 12g \cos 25^{\circ}\)

Substitute \(F\) into the equation for forces parallel to the plane:

\(P + 0.35 \times 12g \cos 25^{\circ} = 12g \sin 25^{\circ}\)

Calculate \(F\):

\(F = 0.35 \times 12 \times 9.8 \times \cos 25^{\circ} \approx 38.1 \text{ N}\)

Substitute \(F\) back into the equation:

\(P + 38.1 = 12 \times 9.8 \times \sin 25^{\circ}\)

\(P + 38.1 = 50.7\)

\(P = 50.7 - 38.1 = 12.6 \text{ N}\)

Thus, the least possible value of \(P\) is \(12.6 \text{ N}\).

To solve for the greatest possible value of \(P\), we resolve the forces parallel and perpendicular to the plane.

1. Resolve forces perpendicular to the plane:

\(12g \cos 25 = R + P \sin 8\)

2. Resolve forces parallel to the plane:

\(P \cos 8 = F + 12g \sin 25\)

3. The frictional force \(F\) is given by:

\(F = 0.3R\)

4. Substitute \(F = 0.3R\) into the parallel force equation:

\(P \cos 8 = 0.3R + 12g \sin 25\)

5. Substitute \(R = 12g \cos 25 - P \sin 8\) from the perpendicular equation into the parallel equation:

\(P \cos 8 = 0.3(12g \cos 25 - P \sin 8) + 12g \sin 25\)

6. Solve for \(P\):

\(P = 80.8\)

Resolve forces perpendicular to the plane:

\(T \sin 60 + R = 25 \cos 20\)

Resolve forces parallel to the plane:

\(T \cos 60 = F + 25 \sin 20\)

\(T \cos 60 + F = 25 \sin 20\)

Using the friction formula \(F = \mu R\):

\(T \cos 60 = 25 \sin 20 + 0.3(25 \cos 20 - T \sin 60)\)

Substitute \(F = 0.3R\) into the equation:

\(T = \frac{25 \sin 20 + 0.3 \times 25 \cos 20}{\cos 60 + 0.3 \sin 60}\)

Calculate the values:

\(T = 6.26\) N (least value)

\(T = 20.5\) N (greatest value)

(i) The normal reaction force \(R\) is given by:

\(R = 3g \cos 60^{\circ}\)

Using the equation for frictional force \(F = \mu R\), we resolve forces parallel to the plane:

\(3g \sin 60^{\circ} - \mu (3g \cos 60^{\circ}) - 15 = 0\)

Solving for \(\mu\), we find:

\(\mu = 0.732\)

(ii) The maximum force is given by:

\(3g \sin 60^{\circ} + F = 3 \sin 60^{\circ} + \mu (3 \cos 60^{\circ})\)

Solving for \(X\), we find:

\(X = 37.0\)

First, calculate the normal reaction \(R\) on the particle:

\(R = 3g \cos 20^{\circ}\)

Next, calculate the frictional force \(F\) using the coefficient of friction \(\mu = 0.35\):

\(F = 0.35 \times 3g \cos 20^{\circ}\)

For the least possible value of \(P\), resolve forces parallel to the plane:

\(P_1 + F = 3g \sin 20^{\circ}\)

Solving for \(P_1\):

\(P_1 = 3g \sin 20^{\circ} - F\)

Substitute the values to find \(P_1 = 0.394\) (as given).

For the greatest possible value of \(P\), resolve forces in the opposite direction:

\(P_2 = F + 3g \sin 20^{\circ}\)

Substitute the values to find \(P_2 = 20.1 \text{ N}\).

First, resolve the forces perpendicular to the plane to find the normal reaction \(R\):

\(R = 20g \cos 60^{\circ} = 100\).

The frictional force \(F\) is given by \(F = \mu R = 100\mu\).

Resolve the forces parallel to the plane for maximum and minimum values of \(P\):

For maximum \(P\), \(P_{\text{max}} = 20g \sin 60^{\circ} + F\).

For minimum \(P\), \(P_{\text{min}} = 20g \sin 60^{\circ} - F\).

Given \(P_{\text{max}} = 2P_{\text{min}}\), substitute the expressions:

\(20g \sin 60^{\circ} + F = 2(20g \sin 60^{\circ} - F)\).

Simplify and solve for \(\mu\):

\(20g \sin 60^{\circ} + 100\mu = 40g \sin 60^{\circ} - 200\mu\).

\(300\mu = 20g \sin 60^{\circ}\).

\(\mu = \frac{20g \sin 60^{\circ}}{300} = \frac{\sqrt{3}}{3} = 0.577\).

Resolve the forces perpendicular to the plane:

\(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\)

Resolve the forces parallel to the plane:

\(P \cos 25^{\circ} = F + 12g \sin 25^{\circ}\)

Since the particle is on the point of moving, the frictional force \(F\) is at its maximum value:

\(F = \mu R = 0.8R\)

Substitute \(F = 0.8R\) into the parallel force equation:

\(P \cos 25^{\circ} = 0.8R + 12g \sin 25^{\circ}\)

Substitute \(R = 12g \cos 25^{\circ} + P \sin 25^{\circ}\) into the equation:

\(P \cos 25^{\circ} = 0.8(12g \cos 25^{\circ} + P \sin 25^{\circ}) + 12g \sin 25^{\circ}\)

Simplify and solve for \(P\):

\(P \cos 25^{\circ} = 9.6g \cos 25^{\circ} + 0.8P \sin 25^{\circ} + 12g \sin 25^{\circ}\)

\(P \cos 25^{\circ} - 0.8P \sin 25^{\circ} = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P(\cos 25^{\circ} - 0.8 \sin 25^{\circ}) = 9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}\)

\(P = \frac{9.6g \cos 25^{\circ} + 12g \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \sin 25^{\circ}}\)

Substitute \(g = 9.8\):

\(P = \frac{9.6 \times 9.8 \times \cos 25^{\circ} + 12 \times 9.8 \times \sin 25^{\circ}}{\cos 25^{\circ} - 0.8 \times \sin 25^{\circ}}\)

\(P = 242\)

(a)(i) Resolve forces parallel to the plane:

\(T \cos 20^{\circ} - 5 - 2g \sin 30^{\circ} = 2 \times 1.2\)

Solving for T:

\(T \cos 20^{\circ} = 5 + 2g \sin 30^{\circ} + 2 \times 1.2\)

\(T = \frac{5 + 2g \sin 30^{\circ} + 2 \times 1.2}{\cos 20^{\circ}}\)

\(T = 18.5 \text{ N}\)

(a)(ii) Resolve forces perpendicular to the plane:

\(R = 2g \cos 30^{\circ} - T \sin 20^{\circ}\)

\(Using F = μR:\)

\(5 = \mu (2g \cos 30^{\circ} - T \sin 20^{\circ})\)

Solving for μ:

\(\mu = \frac{5}{2g \cos 30^{\circ} - T \sin 20^{\circ}}\)

\(\mu = 0.455\)

(b) Calculate the maximum friction force:

\(\text{Max } F = 0.8 \times (2g \cos 30^{\circ} - 15 \sin 20^{\circ})\)

\(= 9.7521 \ldots\)

Net force up the plane:

\(15 \cos 20^{\circ} - 2g \sin 30^{\circ} = 4.0953 \ldots\)

Since 4.0953 < 9.7521, the block does not move up the plane.

(i) Resolve forces down the plane:

\(20 + 5g \, \sin 10^\circ - F = 0\)

\(R = 5g \, \cos 10^\circ\)

\(\mu = \frac{F}{R} = \frac{20 + 8.6824}{49.24}\)

The coefficient of friction \(\mu\) is 0.582.

(ii) Using Newton's second law:

\(5g \, \sin 10^\circ - 0.582 \times 49.24 = 5a\)

Using \(v^2 = u^2 + 2as\):

\(0 = 2.5^2 - 2 \times 4s\)

The distance moved \(s\) is 0.781 m.

Alternative method for part (ii):

Potential energy loss:

\(5g \, d \, \sin 10^\circ\)

\(\frac{1}{2} \times 5 \times 2.5^2 + 5g \, d \, \sin 10^\circ = 0.582 \times 5g \, d \, \cos 10^\circ\)

The distance moved \(d\) is 0.781 m.

First, calculate the loss in kinetic energy (KE):

\(\text{KE loss} = \frac{1}{2} \times 12000 \times (24^2 - 16^2) = 3000000 \text{ J}\).

Next, calculate the gain in potential energy (PE):

\(\text{PE gain} = 12000 \times 9.8 \times 25 = 2940000 \text{ J}\).

The work done (WD) against resistance is:

\(\text{WD against resistance} = 7500 \times 500 = 3750000 \text{ J}\).

The work done by the driving force (DF) is given by:

\(\text{WD by DF} = \text{PE gain} - \text{KE loss} + \text{WD against resistance}\).

\(\text{WD by DF} = 2940000 - 3000000 + 3750000 = 3690000 \text{ J}\).

The driving force is then:

\(\text{Driving force} = \frac{3690000}{500} = 9660 \text{ N}\).

Using Newton's second law, the equation of motion for the box is:

\(P - 8g \sin 5^{\circ} - F = 8a\)

For \(P = 7X\) and acceleration \(a = 0.15 \text{ m/s}^2\):

\(7X - 8g \sin 5^{\circ} - F = 8 \times 0.15\)

For \(P = 8X\) and acceleration \(a = 1.15 \text{ m/s}^2\):

\(8X - 8g \sin 5^{\circ} - F = 8 \times 1.15\)

Solving these equations simultaneously gives:

\(X = 8\)

Substitute \(X = 8\) into one of the equations to find \(F\):

\(F = 56 - 8g \sin 5^{\circ} - 8 \times 0.15\)

\(F = 47.8 \text{ N}\)

The normal reaction \(R\) is given by:

\(R = 8g \cos 5^{\circ}\)

\(R = 79.695 \text{ N}\)

The coefficient of friction \(\mu\) is:

\(\mu = \frac{F}{R} = \frac{47.8}{79.7}\)

\(\mu = 0.600\)

First, calculate the change in potential energy (PE) as the block moves from A to B:

\(\text{PE change} = 60g \times 17.5 = 10500 \text{ J}\)

Next, calculate the change in kinetic energy (KE):

\(\text{KE change} = \frac{1}{2} \times 60 \times (8.5^2 - 3.5^2) = 1800 \text{ J}\)

The work done (WD) against resistance is:

\(\text{WD against resistance} = 6 \times 250 = 1500 \text{ J}\)

The total work done by the pulling force is the sum of the changes in PE, KE, and the work done against resistance:

\(\text{WD by pulling force} = 10500 + 1800 + 1500 = 13800 \text{ J}\)

Using the formula for work done by the pulling force:

\(\text{WD by pulling force} = 50 \cos \alpha \times 250\)

Equating the two expressions for work done:

\(50 \cos \alpha \times 250 = 13800\)

Solve for \(\cos \alpha\):

\(\cos \alpha = \frac{13800}{12500} = \frac{102}{125}\)

Finally, calculate \(\alpha\):

\(\alpha = \cos^{-1} \left( \frac{102}{125} \right) \approx 35.3^\circ\)

(i) Using Newton's second law and the equation for friction \(F = \mu R\), we have:

\(-\left(\frac{1}{3}\right)(W \cos \alpha) - W \sin \alpha = \left(\frac{W}{g}\right)a\)

Substituting \(\sin \alpha = 0.28\) and \(\cos \alpha = \sqrt{1 - \sin^2 \alpha} \approx 0.96\), we get:

\(-(0.32 - 0.28)g = a\)

\(a = -6\).

(ii) Using the equation \(0 = u^2 + 2as\) where \(u = 5.4 \text{ m s}^{-1}\) and \(a = -6 \text{ m s}^{-2}\):

\(0 = (5.4)^2 + 2(-6)s\)

Solving for \(s\), we find:

\(s = 2.43 \text{ m}\)

(i) Apply Newton's second law to the cyclist and bicycle system. The net force equation along the incline is:

\(F - 780 \times \frac{36}{325} - 32 = 78 \times (-0.2)\)

Solving for \(F\):

\(F = 78 \times (-0.2) + 780 \times \frac{36}{325} + 32\)

\(F = -15.6 + 86.4 + 32\)

\(F = 102.8\)

Rounding to the nearest whole number, \(F = 103\) N.

(ii) Use the equation of motion:

\(0 = 7^2 + 2(-0.2)s\)

\(0 = 49 - 0.4s\)

\(0.4s = 49\)

\(s = \frac{49}{0.4}\)

\(s = 122.5 \text{ m}\)

(i) Resolve forces parallel to the plane: The force equation is \(F = 5.9 - 6.1 \sin \alpha\). The normal reaction is \(R = 6.1 \cos \alpha\). For equilibrium, \(5.9 - 6.1 \sin \alpha \leq \mu (6.1 \cos \alpha)\). Given \(\tan \alpha = \frac{11}{60}\), \(\sin \alpha = \frac{11}{61}\) and \(\cos \alpha = \frac{60}{61}\). Substituting, \(5.9 - 6.1 \times \frac{11}{61} \leq \mu \times 6.1 \times \frac{60}{61}\). Simplifying gives \(\mu \geq \frac{4}{5}\).

(ii) When the force acts down the plane, the net force is \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} > 0\). Simplifying gives \(\mu < \frac{7}{6}\).

(iii) Using Newton's second law, \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} = 6.1 \times 1.7\). Solving gives \(\mu = 0.994\).