Let's solve each question step by step: --- ## **Question 18** **Problem Restatement:** - A 4 kg block falls from rest through .8 m. The kinetic energy (KE) of the system (both blocks) when it has fallen is \( K \). - What will the KE be if the blocks are switched and the 2 kg block falls .8 m instead? ### **Step 1: Find KE for the original system (4 kg block falls)** #### **Initial Potential Energy Loss:** Only the difference in mass provides net force (using Atwood’s machine logic). Potential Energy lost by 4 kg block: \[ PE_{loss} = mgh = 4 \times 9.8 \times .8 = 31.36 \text{ J} \] Potential Energy gained by 2 kg block (it rises): \[ PE_{gain} = mgh = 2 \times 9.8 \times .8 = 15.68 \text{ J} \] Net loss in potential energy for the system: \[ \Delta PE = (4 \times 9.8 \times .8) - (2 \times 9.8 \times .8) = 2 \times 9.8 \times .8 = 15.68 \text{ J} \] By conservation of energy, this is the total kinetic energy \( K \): \[ K = 15.68 \text{ J} \] ### **Step 2: Switch the blocks (2 kg falls, 4 kg rises)** \[ \Delta PE = (2 \times 9.8 \times .8) - (4 \times 9.8 \times .8) = -2 \times 9.8 \times .8 = -15.68 \text{ J} \] But since we want the kinetic energy (magnitude): \[ K' = |2 \times 9.8 \times .8| = 15.68 \text{ J} \] But the question wants the answer in terms of \( K \): **Compare:** - Original: \( K = 15.68 \) - Switched: \( K' = 7.84 \) [Wait, let's check this carefully.] Wait: If you carefully analyze, when the 2 kg block falls, the net drop in gravitational energy is: \[ \Delta PE = (2 - 4) \times 9.8 \times .8 = -2 \times 9.8 \times .8 = -15.68 \text{ J} \] But kinetic energy can't be negative, the system can't move if the heavier block is on the other side (since the 4 kg will never rise by itself if 2 kg is falling). So the only way for the 2 kg block to fall is if it were initially above, so the calculation is valid. But the **magnitude** of KE is the same as before, but the system moves in the opposite direction. But let's check the answer choices: - \( a. 4K \) - \( b. .25K \) - \( c. .5K \) - \( d. K \) - \( e. 2K \) But if you check, the net potential energy change is proportional to the mass **difference**: - For 4 kg falling: \( (4-2) \) - For 2 kg falling: \( (2-4) \) But the **kinetic energy** will be the same in magnitude because it's always the mass difference times \( g h \). So the answer is: **Final Answer for 18:** \[ \boxed{K} \] Which is option **d**. --- ## **Question 19** **Given:** - .5 kg mass rotates in a circle of radius .2 m. - 2.5 kg mass hangs vertically, connected by a string. - The system is in equilibrium (the hanging mass is not moving). **Find:** The required rotational rate (revolutions per second) for the \(.5\) kg mass. ### **Step 1: Forces on the Hanging Mass** Tension \( T \) in the string must balance the weight: \[ T = Mg = 2.5 \times 9.8 = 24.5 \, \text{N} \] ### **Step 2: Forces on the Rotating Mass** The tension provides the centripetal force: \[ T = m \omega^2 r \] \[ 24.5 = .5 \times \omega^2 \times .2 \] \[ 24.5 = .1 \omega^2 \] \[ \omega^2 = \frac{24.5}{.1} = 245 \] \[ \omega = \sqrt{245} \approx 15.65 \, \text{rad/s} \] ### **Step 3: Convert \(\omega\) to revolutions per second** \[ 1 \text{ rev} = 2\pi \text{ rad} \] \[ \text{revs/sec} = \frac{\omega}{2\pi} = \frac{15.65}{6.283} \approx 2.49 \text{ rev/s} \] ### **Step 4: Choose Closest Answer** - a. 1 rev/s - **b. 2.5 rev/s** - c. 5 rev/s - d. 7.5 rev/s - e. 10 rev/s **Final Answer for 19:** \[ \boxed{2.5 \text{ rev/s}} \] --- ## **Summary of Answers** **18.** d. \( K \) **19.** b. \( 2.5 \text{ rev/s} \)