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.
ScanSolve AI Answer
Here's a step-by-step solution to the problem: Given Information: Mass of the object, m = 24 \, kg Angle of inclination, = 37^ Initial velocity, v_0 = 0 \, m/s (starts from rest) Distance covered, d = 18 \, m Time taken, t = 3.0 \, s Acceleration due to gravity, g = 9.8 \, m/s^2 The plane is frictionless. --- Step 1: Calculate the acceleration (a) of the object. Since the object starts from rest and moves a certain distance in a given time, we can use the kinematic equation: d = v_0 t + (1)/(2)at^2 Given v_0 = 0: d = (1)/(2)at^2 18 \, m = (1)/(2) a (3.0 \, s)^2 18 = (1)/(2) a (9.0) 18 = 4.5a a = (18)/(4.5) a = 4.0 \, m/s^2 Step 2: Calculate the applied force (F_applied) acting on the object. The forces acting along the incline are the applied force (F_applied) upwards and the component of gravity (mg ) downwards. According to Newton's second law, the net force is F_net = ma. F_net = F_applied - mg ma = F_applied - mg F_applied = ma + mg F_applied = m(a + g ) Using (37^) ≈ 0.6018: F_applied = 24 \, kg (4.0 \, m/s^2 + 9.8 \, m/s^2 × (37^)) F_applied = 24 (4.0 + 9.8 × 0.6018) F_applied = 24 (4.0 + 5.89764) F_applied = 24 (9.89764) F_applied = 237.54336 \, N --- Part a) What is the average power required to accomplish the process? Step 3: Calculate the total work done (W_total) by the applied force. Work done is the product of force and displacement in the direction of the force: W_total = F_applied · d W_total = 237.54336 \, N × 18 \, m W_total = 4275.78048 \, J Step 4: Calculate the average power (P_avg). Average power is the total work done divided by the time taken: P_avg = W_totalt P_avg = 4275.78048 \, J3.0 \, s P_avg = 1425.26016 \, W Rounding to three significant figures: P_avg ≈ 1430 \, W The average power required is 1430 W. --- Part b) What is the instantaneous power required at the end of the 3.0 second interval? Step 5: Calculate the final velocity (v_f) at t = 3.0 \, s. Using the kinematic equation: v_f = v_0 + at v_f = 0 + (4.0 \, m/s^2)(3.0 \, s) v_f = 12.0 \, m/s Step 6: Calculate the instantaneous power (P_inst) at t = 3.0 \, s. Instantaneous power is the product of the applied force and the instantaneous velocity: P_inst = F_applied · v_f P_inst = 237.54336 \, N × 12.0 \, m/s P_inst = 2850.52032 \, W Rounding to three significant figures: P_inst ≈ 2850 \, W The instantaneous power required at the end of the 3.0 second interval is 2850 W.