<p>(a) The particle moves first along the x axis from the origin to the point (3m, 0) and then parallel to the y axis until it reaches (3m, 6m). It is subject to a force F=cxy i^+dj^, where c = 8.0 N/m^2 and d = 13 N. Calculate the work done by the force.</p>
<p>(b) The particle moves first along the y axis from the origin to the point (0, 6m) and then parallel to the x axis until it reaches (3m,6m).It is subject to a force F=cxy i^+dj^, where c = 8.0 N/m^2 and d = 13 N. Calculate the work done by the force.</p>
<p>for (a), I tried doing F * d (dot product) using <144, 13> for F and <3,6> for d and got 510 J which is apparently wrong.</p>
<p>w= Force time displacement.
split it into respective components x and y.
integrate with respect to x.
so like for a) c<em>x^2/2</em>y + d*y
i don’t really know. i just got the right answer using that hahahaha i’m stuck on b
for a though, i think you have to realize that there is no y in the x direction first so that part of the equation becomes 0</p>
<p>ggfailphysics</p>
<p>i think if it’s cx, you have to do an integral because the force isn’t constant - it varies depending on the position.</p>
<p>so for (a), split it up into 2 parts: from (0,0) to (3,0), and then from (3,0) to (3,6)</p>
<p>for the first part, integrate 8x * dx from 0 to 3. you forget about the y-part of the force vector cuz you only care about the horizontal direction. you get 36.</p>
<p>for the second part, you can just do F * distance without integrating. so you would do 13 * 6 = 78</p>
<p>add these two up and you get 114 Joules.</p>
<p>hopefully i didn’t make any mistakes…</p>
<p>try part b doing the same type of analysis</p>
<p>is your way wrong then, i couldnt even find this one on google haha</p>
<p>…and i only have two more shots left for (a) lol</p>
<p>a should be just 78</p>
<p>oh i missed the “cxy”. if that’s the case then y=0 the whole time travelling from (0,0) to (3,0). thus force = 0 during that time and work = 0 as well.</p>
<p>batman is right, i think it should just be 78.</p>
<p>ya Mech and Batman (a) did turn out to be 78.</p>
<p>Batman, do you still not know how to (b)? i figured it out if you still need it.</p>
<p>Thanks guys!</p>
<p>how do you do b? :O</p>
<p>pretty much same logic as for a, except that for (b) you DO have Wx unlike in (a)</p>
<p>Wy:
first integrate F wrt y from y=0 to y=6 and evaluate at x=0 (since ur going straight up). you can ignore Fx here because it goes to 0. Wy evaluates to 13y eval at y=6 and y=0 and ends up being 78 J.</p>
<p>Wx:
integrate F wrt to x from x=0->3 and eval at y=6 (going across horizontally). Ignore Fy here because no work is done in the y direction. this comes out to integrating (8)(6)x from x=0->3, which is 24x^2 eval at x=3 => 216 J</p>
<p>For WT=Wx+Wy=290 J</p>
<p>imma baller, lol</p>
<p>any idea on 9.70 then?</p>