Recent content by racecar12

Oh man, you are correct. I made an algebra mistake while moving everything over to solve for v. Good catch. And thanks a bunch!

Hey physics problems can be tough to get thru sometimes. You can use an equation that relates velocity, time, and distance to get your answer. Remember, Average velocity = Δx divided by Δt...where x is displacement and t is time. The trick here is also that you have two people moving at the...

Hey, thanks for your constructive contribution to my inquiry. I attempted the solution in the way you suggested. I get v to equal 6.3228, which is unfortunately not the correct answer. I'm wondering if I may have to consider the total length the object travels...like adding Δy to the...

Seriously?

Yes, you are correct that kinetic energy equals 0.5mv2. However, removing d from the equation takes me much farther from the answer, which I know to be 5.95617m/s. if I proceed with the way that I have thought, I get v to equal 5.96046. Close, but no cigar. I am making a mistake somewhere...

Thanks for looking at this post. Here was my though process as I approached this problem... We'll just use the conservation of energy theorem for this problem. Einitial=Efinal Kinitial+Uinitial=Kfinal+Ufinal Since the massive object has no initial kinetic energy at the start of the system...

Homework Statement A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 . What is...

I clearly don't understand a lot about physics, forces, etc. If I'm understanding correctly, Fnet = N-W Is Fnet supposed to equal zero? I understand that W is equal to mg. the force of W then is -8.82kN. Is that right? I mean do I account for the downward direction of gravity in this...

I tried it again... There are 2forces acting in this problem. One is the weight force pushing down, which is W=mg. The second force is the normal force pushing up from the floor of the elevator, N=ma. W=mg m=0.9kN g= -9.8 W=-8.82N N=ma m=0.9kN a=-1 N=-0.9N Since the elevator...

Homework Statement Person weighing 0.9kN rides in an elevator that has downward acceleration of 1m/s^2. What is the magnitude of the force of the elevator floor on the person. Answer in units of kN. Homework Equations F=ma The Attempt at a Solution I drew a force body diagram...

Did what you said... Fx = 9.4 Fy = 9.1 F = 13.1N Thank you

Thanks for the clarification about gravity :) I went back to find the y component of the vector...to equal 1.33sin41? I thought 9.4 was the x component of the Force. Does 1.33sin41 equal the y comonent of velocity? If so, then I could go about finiding the y component of acceleration and...

To find Fy, I used the mass of the object x the negative acceleration due to gravity. F= 12.1(-9.8) I'm missing the normal force, I think? But now, I am unsure how to calculate it

Homework Statement Person pushes down the length of the handle of a 12.1kg lawn spreader. The handle makes an angle 44.1 degrees with the horizontal. Person wishes to accelerate the spreader from rest to 1.33m/s in 1.7s. What forece must person apply to the handle?Homework Equations F=ma...

Oh yay we got the same answer :) Thanks a ton. I'm not very good at these types of problems yet...but I will keep practicing.