Recent content by DJW

Is it true that they both must also have zero kinetic energy?

I subtracted that from 250 x cos(26) = 224.7 and got 44.8. Then I did F=ma which is 44.8=50(a) and got a=.896 (multiple choice answer is .894, which is the correct answer.). I had the normal force all messed up. Thanks for the help.

Fk=μkN, right? (.30 x 599.59=179.877)?

Oh, right. I already had the sum of the Y forces, but I guess I wasn't paying attention. I did mg + sin(26) x 250 = 599.59. Do I find ΣFx, add the two components, then solve with F=ma?

Hmm...250N, mg, and N? Not sure about the third one.

Would that be mgcosΘ?

Hey, I was needing some help with a friction problem involving finding the acceleration of an object. A 50.0-kg box is being pushed along a horizontal surface by a force of 250 N directed 26 degrees below the horizontal. The coefficient of kinetic friction between the box and the surface is...

Thanks.

Does that mean that the velocity of a projectile at the highest point of its trajectory will be whatever the initial X velocity is?

Hey, I had a basic conception question about projectile velocity. When a projectile reaches its highest point, does the velocity of the projectile equal 0 or does it equal the initial x velocity? I know that Vy=0 at the highest point of a projectile's trajectory, but I'm not sure about the other...