New Users New to PHF? Post up here and introduce yourself! Oct 22nd 2018, 08:20 PM #1 Junior Member   Join Date: Jun 2018 Posts: 7 Question about Force, Motion and Friction I have a problem with this question: A skip of mass 2 tonnes has to be dragged along a track. The coefficient of friction is 0.7. A cable attached to the skip runs horizontally to a winch powered by an electric motor. Using the controls on the motor, the tension in the cable is gradually increased, so that after t seconds its value is 400t newtons, until it reaches its maximum value of 16 000 newtons. This maximum tension is then reduced instananeously so that the skip continues to move at this speed. a) How long does it take before the skip starts to move? 400t = 0.7*20 000. So t = 35 seconds. b) How fast is the skip moving when the tenstion in the cable reaches its greatest value? My working for b) above would be: 400t = 16 000. t = 40 So it reaches its maximum value after 40 seconds, which means 40 - 35 = 5 seconds after the skip stars to move. Net force = Force - Friction = 16 000 - 0.7 X 20 000 = 16 000 - 14 000 = 2000 N = ma = 2000a So a (acceleration) = 1 ms^(-2) Thus, v = u + at = 0 + 1*5 = 5 ms^(-1) However, the text says that the answer is 2.5 ms^(-1), which is half of what I have ended up with. Please tell me what is wrong with my working and what is missing in it (if any). Thank you.   Oct 23rd 2018, 05:02 AM #2 Senior Member   Join Date: Jun 2016 Location: England Posts: 881 You have applied the final cable tension of 16000N over the entire 5 seconds. however it is varying by 400N/s __________________ ~\o/~ Last edited by Woody; Oct 23rd 2018 at 05:05 AM.   Oct 23rd 2018, 09:21 PM #3 Junior Member   Join Date: Jun 2018 Posts: 7 Thank you for your help. I worked it out as follows: F = 400t = ma = 2000a So t = 5a So a = (1/5)t You intergrate it with respect to t to get: v =(1/10)t^2 + C v = 0 when t = 0 so C = 0 So v = (1/10)t^2 Thus, when t = 5, v = (1/10)*5^2 = 25/10 = 2.5 ms^(-1)   Nov 7th 2018, 06:58 AM #4 Senior Member   Join Date: Aug 2010 Posts: 404 You have just one "coefficient of friction". Are we to use that for both "static friction" and "sliding friction"? Also when the skip starts to move, at 35 seconds, the tension in the cable is 14000 N. AT 40 seconds, it is 16000 N. Because it increases linearly you can use the average value 15000 N for the entire 5 seconds.   Nov 7th 2018, 03:55 PM #5 Junior Member   Join Date: Jun 2018 Posts: 7 I do not know the difference between "static" friction and "sliding" friction as I am a beginner of Physics. What is it? But I know that, as long as I understand from this textbook, this coefficient of friction applies to the friction acting in the opposite direction of the movement of the object. So I guess that you can use it for "sliding" friction. Also I am not sure what you really want to imply when you say "at 35 seconds, the tension in the cable is 14000 N. AT 40 seconds, it is 16000 N. Because it increases linearly you can use the average value 15000 N for the entire 5 seconds". But can we use the momentum formula for this question? As you say, the difference in tensions is 16000 - 14000 = 2000 So the average would be 2000/2 = 1000 Then use the formula Ft = mv - mu where F = 1000, t = 5, m = 2000, u = 0, So 1000*5 = 2000v - 2000*0 v = 5000/2000 = 2.5 ms^(-1) Is that what you wanted to imply or something else?  Tags force, friction, motion, question, velocity or acceleration Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post osalselaka Kinematics and Dynamics 1 Jun 11th 2017 11:41 AM sbnett General Physics 2 Jun 5th 2017 06:44 PM Dirkpitt45 Kinematics and Dynamics 1 Dec 15th 2009 03:02 PM jonbrutal Kinematics and Dynamics 6 Feb 5th 2009 06:42 AM Morgan82 Kinematics and Dynamics 2 Dec 1st 2008 12:45 AM