This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions for Question 2 from the image you sent. QUESTION TWO a) A 100g body is dug up by a force, moving uniformly through 100m in 10 seconds. Determine: (i) The work done (ii) The power Assumption: The "body" refers to the 100g of earth, and it is lifted vertically (dug up) uniformly through a distance of 100m. Step 1: Identify the given values and calculate the force. Mass m = 100 \, g = 0.1 \, kg. Distance (height) h = 100 \, m. Time t = 10 \, s. Acceleration due to gravity g = 9.8 \, m/s^2. Since the body is moving uniformly, the applied force F is equal to its weight. F = mg F = (0.1 \, kg)(9.8 \, m/s^2) F = 0.98 \, N Step 2: Calculate the work done. Work done W is the force multiplied by the distance moved in the direction of the force. W = Fh W = (0.98 \, N)(100 \, m) W = 98 \, J The work done is 98 J. Step 3: Calculate the power. Power P is the work done divided by the time taken. P = (W)/(t) P = 98 \, J10 \, s P = 9.8 \, W The power is 9.8 W. b) A load is hoisted 4m and requires a force of 500 N. The efficiency of the hoist gear is 80% and that of the motor is 70%. Determine the input energy to the hoist motor. Step 1: Calculate the useful output work (work done on the load). The force required to hoist the load is the output force, and the distance is the height. W_out = F_load × h W_out = 500 \, N × 4 \, m W_out = 2000 \, J Step 2: Calculate the energy input to the hoist gear. The efficiency of the hoist gear (_gear) is the ratio of the work output from the gear to the energy input to the gear. _gear = W_outE_input\_gear E_input\_gear = W_out_gear E_input\_gear = 2000 \, J0.80 E_input\_gear = 2500 \, J Step 3: Calculate the input energy to the hoist motor. The energy input to the hoist gear is the energy output from the motor (E_output\_motor). The efficiency of the motor (_motor) is the ratio of the energy output from the motor to the energy input to the motor. _motor = E_output\_motorE_input\_motor E_input\_motor = E_output\_motor_motor E_input\_motor = 2500 \, J0.70 E_input\_motor ≈ 3571.43 \, J The input energy to the hoist motor is 3571.43 J. That's 2 down. 3 left today — send the next one.