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
Okay, Jordan, let's work through Question 6. QUESTION 6: (10 marks) A bodybuilder pulls a 60 \, kg tyre with a force of 250 \, N at an elevation of 20^ to the horizontal. 1- If he is able to pull for 25 metres, calculate the work done by the bodybuilder. Step 1: Calculate the work done by the bodybuilder. The work done by a force at an angle is given by W = F d . Here, F = 250 \, N, d = 25 \, m, and = 20^. W_B = F d W_B = (250 \, N)(25 \, m) (20^) W_B = 6250 \, J × 0.9397 W_B ≈ 5873.125 \, J 2- Provided that the tyre was initially at rest and there is an additional frictional force of 220 \, N: a) Determine the work required to overcome the frictional force. Step 2: Calculate the work done against friction. The work done against friction is W_f = F_f d. Here, F_f = 220 \, N and d = 25 \, m. W_f = F_f d W_f = (220 \, N)(25 \, m) W_f = 5500 \, J b) Using the kinetic energy theorem, deduce the kinetic energy K_E the bodybuilder provides the tyre. Step 3: Apply the Work-Energy Theorem to find the kinetic energy. The net work done on the tyre is the work done by the bodybuilder minus the work done against friction. This net work equals the change in kinetic energy ( K_E). Since the tyre starts from rest, the initial kinetic energy is 0. W_net = W_B - W_f W_net = 5873.125 \, J - 5500 \, J W_net = 373.125 \, J So, the kinetic energy provided to the tyre is: K_E = 373.125 \, J c) Calculate the tyre's velocity. Take K_E = 373 \, J. Step 4: Calculate the velocity using the kinetic energy formula. The kinetic energy is given by K_E = (1)/(2)mv^2. We are given K_E = 373 \, J and m = 60 \, kg. K_E = (1)/(2)mv^2 373 \, J = (1)/(2)(60 \, kg)v^2 373 \, J = (30 \, kg)v^2 v^2 = 373 \, J30 \, kg v^2 = 12.433 \, m^2/s^2 v = sqrt(12.433 \, m)^2/s^2 v ≈ 3.526 \, m/s What's next?