February 2nd, 2015, 01:45 AM  #1 
Senior Member Joined: Sep 2012 Posts: 201 Thanks: 1  work done up slope question
Hi I am have a bit of trouble with this questions as the ans in the book tell me a diffrent ans could someone please checking my working to see where I am going wrong. Question: Find the work done in pulling a packing case of mass 80kg a distance of 15m against a constant resistance of 150N. B) up an incline whose angle $\displaystyle \theta$ to the horizontal is given by $\displaystyle sin\theta=\frac{1}{8}$ My working: Work done agaist friction: $\displaystyle 150\times 15=2250J$ Work done agaist gravity: $\displaystyle 80g\times 15sin\theta=80g\times 15\frac{1}{8}=1470J$ $\displaystyle \therefore$total wd=$\displaystyle 3720$ Ans in book $\displaystyle 4350J$ Big thanks in advance p.s $\displaystyle g=9.8$ well thats what I took it as. 
February 2nd, 2015, 02:49 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,119 Thanks: 710 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I get the same answer as you.

February 2nd, 2015, 03:59 AM  #3 
Senior Member Joined: Sep 2012 Posts: 201 Thanks: 1 
See I dont know if a typo in the book or I have complety missunderstood the question. How did you calculate it?

February 2nd, 2015, 06:03 AM  #4 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,119 Thanks: 710 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
To push the crate up the wall, your force must exceed $\displaystyle F = F_{resistance} + mg\sin\theta$ $\displaystyle = 150 + 80\times 9.81 \times \frac{1}{8}$ $\displaystyle = 248.1$ Therefore, the minimum work done is $\displaystyle WD_{min} = F\cdot d = 248.1\times 15 = 3721.5 J$ You can also calculate it by adding the work done against resistance by the rise in GPE: $\displaystyle WD_{min} = F_{resistance}\cdot d + mg\Delta h$ $\displaystyle = 150\times 15 + 80\times 9.81 \times (15\sin\theta)$ $\displaystyle = 2250 + 80\times 9.81 \times 1.875$ $\displaystyle = 2250 + 1471.5$ $\displaystyle =3721.5$ This is a minimum work done. Typically there will be heat losses and the requirement to impart kinetic energy to the mass, so the force will have to exceed the minimum force and work done must be higher. Perhaps there is extra information in the question relating to the kinetic energy obtained by the mass? 

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