Assign 7 Begin Date: 10/18/2017 9:00:00 AM Due Date: 10 23/2017 9:00:00 AM End D
ID: 3281019 • Letter: A
Question
Assign 7 Begin Date: 10/18/2017 9:00:00 AM Due Date: 10 23/2017 9:00:00 AM End Date: 10/25/2017 9:00:00 AM (20%) Problem 1: You want to move your m-29 kg spare tire from the floor onto the back ofyour pickup truck. The bed of your truck is h-0.91 m above the floor. You have found a sturdy rope, and a strong plank to use as a ramp that is d-3.3 m long. The coefficient of kinetic friction between the tire and the plank is . Case 2 Case 3 ©theexpertta.com 14% Part (a) What is the minimum work in Joules you will do in rolling the tire all the way up the ramp as shown in Case (l)? any losses due to friction. eglect 14% Part (b) Calculate the size of the average force in Newtons that you must apply on the tire if you push in a dnection parallel to the ramp? See Case (1). - 14% Part (c) After a few unsuccessful tries your friend suggests that you push the tire up the ramp by applying a horizontal force, as shown in Case (2). Calculate the work in Joules you will (hopefully) do in moving the tire up the ramp this way. Again, neglect any frictional losses. 14% Part (d) Calculate the size of the average force in Newtons that yo u must apply on the tire ifyou push horizontally to roll it up the ramp. See Case (2) 14% Part (e) In yet another desperate attempt, you think perhaps that it will be easier to pull the tire up the ramp. You put the tire on the ramp, tie your sturdy rope to the tire, and pull on the rope parallel to the ramp, as shown in Case 3. Calculate the work in Joules you will (hopefully) do in moving the tire up the ramp this way given -0.335 14% Part (f) Considering Case l which of the following could you do to decrease the amount of work you do in moving the tire from the floor onto the bed of your pickup truck? Select the best response.Explanation / Answer
a] Minimum work = mgh = 29*9.8*0.91 = 258.6 J
b] Work =mgh = Fd
F = mg h/d = 258.6/3.3
= 78.36 N answer
c] work = Fd cos theta = 258.6 J
d] cos theta = cos arcsin(0.91/3.3) = 0.961
F = 258.6/(3.3*0.961) = 81.54 N answer