Cotrolled Test 1 Grade 11 A trolley is pushed along a horizontal surface with a force of 150 at angle of \( 45^{\circ} \) to the horizontal. The trolley experiences a consta frictional force of 60 N . The net force on the trolley... (i) causes the trolley to accelerate horizontally. (ii) is equal to applied force. \( \begin{array}{l}\text { (iii) is horizontally forward. } \\ \text { Which of the statements above are CORRECT? } \\ \text { A (i) and (ii) } \\ \text { C } \\ \text { (i) and (iii) } \\ \text { Mary has a mass of } 50 \text { kg and is standing on a scale in a lift. Th } \\ \text { reading on the scale is } 555 \mathrm{~N} \text {. The lift is... } \\ \text { A accelerating upwards. } \\ \text { B moving downwards at constant velocity. } \\ \text { (iii) } \\ \text { moving upwards at a constant velocity. } \\ \text { accelerating downwards. }\end{array} \)

1. **Trolley Problem** - **Step 1. Decompose the applied force** The force is \(150\,\mathrm{N}\) at an angle of \(45^\circ\). Its horizontal and vertical components are: \[ F_{\text{horizontal}} = 150\cos 45^\circ = 150\left(\frac{\sqrt{2}}{2}\right) = 75\sqrt{2}\,\mathrm{N} \approx 106.1\,\mathrm{N} \] \[ F_{\text{vertical}} = 150\sin 45^\circ = 150\left(\frac{\sqrt{2}}{2}\right) = 75\sqrt{2}\,\mathrm{N} \approx 106.1\,\mathrm{N} \] - **Step 2. Determine the net horizontal force** Friction opposes the motion with a force of \(60\,\mathrm{N}\). The net horizontal force is: \[ F_{\text{net,horizontal}} = F_{\text{horizontal}} - 60\,\mathrm{N} \approx 106.1\,\mathrm{N} - 60\,\mathrm{N} = 46.1\,\mathrm{N} \] Since the net horizontal force is nonzero and directed forward, the net force: - **(i)** Causes the trolley to accelerate horizontally. - **(iii)** Is horizontally forward. Statement **(ii)** is incorrect because the net force is not equal to the applied force; friction reduces it. - **Conclusion for the trolley problem:** Statements **(i)** and **(iii)** are correct. 2. **Lift Problem (Mary's Case)** - **Step 1. Compute Mary’s weight** Mary's mass is \(50\,\mathrm{kg}\). The weight is given by: \[ mg = 50\,\mathrm{kg} \times 9.8\,\mathrm{m/s^2} = 490\,\mathrm{N} \] - **Step 2. Use the scale reading to find the acceleration** The scale reading (normal force) is \(555\,\mathrm{N}\). When a person is in an accelerating lift, the normal force \(N\) is given by: \[ N = m(g+a) \] Solve for the acceleration \(a\): \[ a = \frac{N}{m} - g = \frac{555\,\mathrm{N}}{50\,\mathrm{kg}} - 9.8\,\mathrm{m/s^2} = 11.1\,\mathrm{m/s^2} - 9.8\,\mathrm{m/s^2} = 1.3\,\mathrm{m/s^2} \] Since the acceleration is upward, the lift is accelerating upwards. - **Conclusion for the lift problem:** The lift is accelerating upwards. **Final Answers:** - For the trolley, statements **(i)** and **(iii)** are correct. - For Mary in the lift, the lift is accelerating upwards.