Recent content by scotthyppo

What do you mean by that? So the magnitude of the acceleration should just be 6.063

So -mg sin(30) - mew = max -mg sin (30) - .137 mg cos (30) = max So the m cancels on both side -g sin (30) - .137 g cos (30) = ax acceleration = -6.063 For the total work done by gravity on the box W = F (cos (30)) V^2 = Vo^2 + 2ax 0 = 5^2 + 2 (-6.063)x = 25 + (-12.126) x X= 2.06 W =...

Yep This is what I did for my calculation: Fy = N - mg cos (30) = 0 N = mg cos (30) N = 42.4352 Fx I did: (42.4352x .137) x (5x9.8) x sin(30) and got 142.434 x direction you should have two forces: the force due to friction and the weight of the object that is parallel to the surface How do...

No he only declining with more information

I am still confused can you show me what you mean?

For my X component I got a 55.3136

Fy = 0 mg cos (30) Fx = ma + coefficient + mg sin (30) What do you do after this?

Ive tried finding the distance but could not

Can someone help me A 5kg box slides up an inclined plane surface, with an initial speed of 5.00 m/s. The plane makes a 30 degrees angle with horizontal. The coefficient of friction is .137. Assume g = 9.80 m/s^2. A) what is the magnitude of the acceleration of the box? B) What is the...