Recent content by spacestrudel

Got’cha! I don’t know, haha I was just taking the stuff he wrote on the board literally. Okay! So take the integral of the function given to then use to find the impulse ! Thank you.

Does it still have to be the integral of that given function even though it is already in terms of F(t)? Because he said that the change in momentum was just the integral of F. So, since I have F(t) - isn't that technically the integral?

Hi there, Just asking a logistics question since I want to be sure I am approaching this problem correctly. My professor showed me an example of a bullet being fired from a barrel, given its initial velocity was 0. The change in time was 0.1 seconds. The mass of the bullet is 0.02 kg. The...

Okay ! I see - thank you. I used the following equation for calculating the final velocity: v^2i +2gxsinΘ -2ukxcos(Θ) = v^2f The 2 comes from the fact that I multiplied everything by 2 to get rid of the 1/2's. Also since everything had "m" on it, I also canceled out all of the m's. Then I...

Ah, I see. Here is my work: 1/2mv^2f +mghf = ukmgcos(Θ)(x+s) + 1/2ks^2 By plugging in the values I have the following: 1/2(300)(10.727)^2 + (300)(9.81)[-((s+5)sin(30))] = (0.4)(300)(9.81)cos(30)(s+5) + 10s^2 (where 300 = the mass, 10.72 = the final velocity, 9.81 for gravity, sin(30) = 1/2...

Thank you! I received a rather small number for the max. compression (1.914 meters) of the spring, but I'm assuming this makes sense since the spring constant was only 20 N/M and the block had only traveled a bit more than 5 meters at this point. The final velocity of the block that I calculated...

I believe it stems from the fact that I should actually say hf = -xsin(30), since it's moving down on an incline? And do I have to add s to the FE side, as well?

Ah, I see. So, I'm assuming it should be vi^2 +2*g*x*sin(30) ...) since gravity would be speeding the block up. Which would then change my value of vf. So, if the spring is compressed by s, could I write that hf = (s+x)sin(30) or something along those lines? (We know that x = 5 meters from the...

Ah, sorry about that I will write out my steps for part a. Also, after redoing it again at a reasonable hour instead of at 2 am, I got a different answer for Vf. So, since part a) asks for the final velocity of the block right before impact, I did the following steps: KEi + ug0 = KEf + ugf +...

So sorry let me clarify - part a) is the scanned image included in this post. The question asks, "what is the impact velocity?" So, by solving using conservation of energy, as well as replacing values such as the fact that FE can be replaced to equal ukFNd which simply replaces to FN = mgcos(Θ)...

The force of friction I found as FE = ukFNd (where uk represents the coefficient of kinetic friction). By plugging in the normal force and x for d, I then found this to be 2ukgcos(Θ)x. (At least from above in terms of the final velocity calculations I performed for part a). And I am given that...

Hello! I am stuck on part of a problem and was wondering what I am doing wrong. For part a of the problem, we were asked to find the impact speed. I did this in a photo below given the following values: Θ = 30 degrees. The initial velocity = 10 m/s. The coefficient of kinetic friction = 0.4...