Recent content by Tim Wellens

Homework Statement If block A slides with a velocity into block B which is at rest, and after this collision block B ends up with a velocity and block A is stationary after. Block B has a mass of 5m. Would this type of collision be an elastic collision or inelastic? How do you know? What...

I was just kind of thinking about it in terms of -1 - 1. The sign of MaVaf is negative, the sign of MaVai is positive. In terms of kinetic energy, both -MaVaf and MaVai would have positive kinetic energy due to the block being in motion. Block B would initially have zero kinetic energy, but...

Homework Statement The whole problem is in variables, which tends to confuse me a little. We have two blocks that collide, Ma and Mb. Initially Ma experiences a positive velocity. Mb experiences a 0 velocity. Finally, Ma experiences a negative velocity, while we do not yet know Mb's final...

I think Y is vertical to the plane and the forces in the y direction equal 0, while the forces in the x direction equal mgsinΘ.So, the net force is mgsinΘ. Is it correct that the net force would remain constant as the block travels from point A to B because the mass, gravity, and angle stay...

Oh, ok. So, the normal force will be zero because it's force is perpendicular to the displacement vector, and the weight would be positive because it's force vector makes an angle of less than 90 degrees with the displacement vector. So, we always use the weight without breaking it into...

Homework Statement The question is about a block on an incline at an angle Θ above the horizontal. The block will begin from the resting position and travel from point A to B down the incline. Neglect friction. 1) As the block travels from point A to B, will the magnitude of the net force...

Ok, I'll back up to the previous question that wanted to know how far the spring stretched while it was at rest. I had this: Fspring=mgsinΘ= (20kg)(9.8m/s^2)(sin30)= 98 N Then I used Fspring=kx 98 N= (500N/m) x x= 98/(500 N/m)= 0.20 m But, you're saying it should be mgsinΘ= -Fspring, which...

So for 1) would i do ΣFx=mgsinΘ+Fspring=ma... turn fill in kx for Fspring to be ΣF=mgsinΘ+kx=ma and solve for x? In this case I get an x value of .134 m. I had another question of how far does the spring stretch while it is at rest and I put since it's at rest ΣF=0 and Fspring=mgsinΘ=kx, so...

https://flic.kr/p/A3aCsM for a picture like that, if that spring wasn't hooked to a wall and I was holding it and pulling it up the ramp like the question says, would the force of the spring be pointing back toward the spring since it's stretched? When the spring is stationary, there is not a...

Homework Statement A block is sitting on a ramp with a spring behind it, which we are supposed to imagine we are holding. The mass of the block is 20 kg, the angle of the ramp is 30 degrees, and the spring constant is 500 N/m. 1) You pull so the block is accelerating at 4 m/s^2 up the ramp...

Yes, that helps clear up the reason behind part A a lot, thank you! :smile: For part b, was I totally off on how to figure out the stretch of the spring if the acceleration is 4 m/s^2?

1. Homework Statement We are dealing with a block of 20 kg mass on a ramp that is at an angle of 30 degrees. The spring constant given is 500 N/m. Picture- https://flic.kr/p/A3aCsM The only difference is that that we are supposed to be holding the spring stationary, it isn't attached to a...

Homework Statement Picture- https://flic.kr/p/zi1ays The rubber stopper will be spun at a constant speed in a horizontal circle. The hanging mass is only in contact with the string.[/B] A) For the rubber stopper, state the forces and what they are exerted by and on B) Does the net force have...

Yes. I meant to say that I got 4.71 x 10^25 for f1x and f1y. For f2x I'd have 1.33 x 10^26 because there is only an x component of this force. So for the total force my Fx would be 4.71 x 10^25 (granted that was correct) + 1.33x 10^26 = 1.80 x 10^26 N Fy= 4.71 x 10^25 N And the magnitude of...

Ok, so for F1x i would be cos45= f1x/6.66x10^25 which would equal 4.71 x 10^27. F1y was the same due to the angle being 45 degrees. Is this right? If so, am i then supposed to use the law of cosines to find what the total force is?