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    Danelle's Avatar
    Danelle Posts: 2, Reputation: 1
    New Member
    		  
    #1

    Oct 28, 2007, 04:38 PM
    Tension and work
    A sledge loaded with bricks has a total mass of 18.0kg and is pulled at constant speed by a rope inclined at 20.0 degrees above the horizontal. The sledge moves a distance of 20.0m on the horizontal surface. The coefficient of kinetic friction between the sledge and surface is .500. What is the tension of the rope? How much work is done by the rope on the sledge? What is the mechanical energy lost dure to friction?
    CaptainRich's Avatar
    CaptainRich Posts: 4,492, Reputation: 537
    Cars & Trucks Expert
    		  
    #2

    Oct 28, 2007, 04:46 PM
    That sounds like a homework question.

    https://www.askmehelpdesk.com/other-...board-b-u.html
    CaptainRich's Avatar
    CaptainRich Posts: 4,492, Reputation: 537
    Cars & Trucks Expert
    		  
    #3

    Oct 28, 2007, 04:57 PM
    Danelle disagrees: His answer did not help me AT ALL. THANKS FOR NOTHING!

    That's because we're not here to answer your homework questions!
    Do you want to be reported for an inappropriate post??

    The least you can do is try yourself first!
    terryg752's Avatar
    terryg752 Posts: 197, Reputation: 4
    Junior Member
    		  
    #4

    Oct 30, 2007, 02:09 AM
    Quote Originally Posted by Danelle
    A sledge loaded with bricks has a total mass of 18.0kg and is pulled at constant speed by a rope inclined at 20.0 degrees above the horizontal. The sledge moves a distance of 20.0m on the horizontal surface. The coefficient of kinetic friction between the sledge and surface is .500. What is the tension of the rope? How much work is done by the rope on the sledge? What is the mechanical energy lost dure to friction?
    I will presume that this is NOT a home-work question but for your practice.

    Now, the weight being pulled is (mg) 18g

    Upward component of tension is T sin 20 (verticle component of tension T)

    So Normal Reaction + T sin 20 = 18g (upward force = Downward force)

    Normal Reaction = (18g - T sin 20)

    Hence friction = .5 (18g - T sin 20)

    Horizontal component of Tension T= T cos 20

    Hence T cos 20 = .5 (18g - T sin 20)

    This gives you value of T

    Work done = (Component of force in the direction of motion) times distance
    = (T cos 20) times 20

    This is also the energy lost due to friction. (There would be no work done were there no friction)
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