Recent content by enrijuan

ah ok so then my original equation is correct? as in W= \DeltaKE + PE ? mu(9.81cos20)(5.1)= -.5vi2 + (9.81)(5.1sin20) ? or should I find the Net Force as JesseC said? If I take that route then Fx=f-Fgx right? Fx=mu(m)(g)-m(g)(sin20) Fx=.54(2.1)(9.81cos20)-2.1(9.81sin20)= 3.41 N Fx= 3.41=...

I thought that there would be force due to gravity, normal force and the force due to kinetic friction? and I drew a diagram and thought that the distance was directly below as in the y-axis and the hypotenuse would be the displacement.

Homework Statement You and your friend Peter are putting new shingles on a roof pitched at 20\circ . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.1 m away, asks you for the box of nails. Rather than carry the 2.1 kg box of nails...

wow thank you so much! thanks for the welcome, the µ, the help, and above all the patience!

then, .5(7)(0)-.5(7)(2.3^2)= (.250)(7)(-9.81)d ? then d=1.08 meters ?

oh the end velocity is goin to be zero right? so, 1/2mvf2-1/2mvi2=W 1/2mvf2-1/2mvi2 = (µmg)(d) is this right?

so, work done= (7.00 kg)(2.3 m/s) + 0? then work done =18.515 J then 18.515=(force due to kinetic friction)(distance)?

doesnt the conservation of mechanical energy talk about potential and kinetic energy? I don't understand how I would get potential energy or how work would help me in the problem.

I know that force over distance is work but I don't see how that would help me in this problem... I am sorry I am having a hard time with this one... I also know that d= vt

Homework Statement A baggage handler drops your 7.00 kg suitcase onto a conveyor belt running at 2.30 m/s. The materials are such that µS= 0.410 and µK= 0.250. Homework Equations How far is your suitcase dragged before it is riding smoothly on the belt? The Attempt at a Solution...