Recent content by iJamJL

Homework Statement A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by: x = 0.484t^3 - 33.6t Find W, the work done by this force during the first 1.49 s. Homework Equations W=F*s W=mgs Integration The Attempt at a...

webassign.net That's what my school uses for assigning homework. I've entered -93.95. Should it be positive? I don't think so, right? What I mean is that maybe Vi and Vf should be: KE=1/2*m*(-(Vf^2)-Vi^2) KE=1/2*8.87*(-(4.49^2)-6.43^2) KE= (-272.77) J Vf is negative because it goes...

You say that my part (a) equation is correct, but what about the numerical value that I get at the end? I'm only asking because my online homework system tells me that it's wrong. I do..sort of understand why we use that equation though.

Fixed part (a). Noticed a huge mistake. EDIT: Part (a) is still wrong..

Hi Mathoholic, It looks like I did part (a) incorrectly. Would it be that for initial velocity, I use 6.43 m/s, and for final velocity, I use 4.49? It would end up like this: KE=1/2*m*(Vf^2-Vi^2) KE=1/2*8.87*(4.49^2-6.43^2) KE= (-93.95) J (b) I don't exactly follow that formula, or why you're...

Homework Statement An incline makes an angle of 29.9° with the horizontal. A 8.87 kg block is given a push up this incline and released. It starts at the bottom with initial speed 6.43 m/s, travels up the incline, stops, and slides back to the bottom at final speed 4.49 m/s. Using energy...

Wow, I'm bad at this. I initially thought that v-initial would be 0, but then I looked back and realized it's not. *facepalm* v(0)=(-33.6)m/s v(1.49)=(-30.4)m/s W=1/2*m*(v2^2) - 1/2*m*(v1^2) W=1/2*6.26(-30.4^2) - 1/2*6.26*(-33.6^2) W=2892.6 - 3533.6 W=(-641) J Did I finally get the correct...

lol I wrote that on my scrap paper, but typed and calculated it incorrectly! My end result came to: v=(.484)*3*(1.49^2) - 33.6 =(-30.4)m/s W=1/2*m*v^2 W=1/2*6.26*(-30.4^2) W=2892.6 J That's still coming out as incorrect.

Thanks for the replies! I didn't realize that until you pointed it out! I tried it out, but I think I made a mistake somewhere. Here's what I did: To find the final velocity, which for this case is at 1.49s, we take the first derivative at 1.49: x = 0.484t3 - 33.6t dx/dt= 2*(.484)t^2 -...

Homework Statement A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by: x = 0.484t3 - 33.6t Find W, the work done by this force during the first 1.49 s. Homework Equations F=mg F=ma W=F*distanceThe Attempt at a Solution To find the...

Okay, so I realized that since this is an incline, we need to include it with the forces: Going up, F(truck) - Friction - mgsin(theta) = 0 That means going down, -(F(truck) + mgsin(theta)) - friction = 0 I don't know whether to add them or set them equal to each other..

Based on what you're saying, it sounds like it goes down the hill at the same speed (19 m/s) as it does going up. However, since we're going downhill, wouldn't we need to account for another factor? I'm not sure what exactly that is since: mgsinθ - Ffriction = ma = 0 mgsinθ = Ffriction...

Oops :roll: lol You're right. I think it was just a bad assumption on my part rather than my professor's fault. Nevertheless, I thank you both for your time and help. :)

Homework Statement A highway goes up a hill, rising at a constant rate of 1.00 m for every 48 m along the road. A truck climbs this hill at constant speed vup = 19 m/s, against a resisting force (friction) f equal to 1/24 of the weight of the truck. Now the truck comes down the same hill...

EDIT: Sorry, the problem was actually the negative. I didn't realize that direction mattered that much! I thought if we had switched it around so that we made gravity in the positive direction, then the amount of work is the same numerical answer, and we could ignore the negative. Big mistake on...