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November 21, 2014

November 21, 2014

Posted by **mary** on Sunday, June 7, 2009 at 10:29pm.

I did W=fdcos(theta) and found the amount of work to be 13.91J.

I then thought maybe you use the equation K=(1/2)mv^2

and rewrite as W=(1/2)mv^2

in which case v=3.12m/s

but this was wrong.

any guidance is appreciated! THANKS

- physics -
**MathMate**, Sunday, June 7, 2009 at 11:40pmm=2.86 kg

g=9.81 m/s/s

θ=30 degrees

F=22 N horizontal

d=0.73 m (displacement)

μ=0 (coefficient of friction)

The weight of the mass can be resolved into two components,

Normal to incline = mg cos(θ)

along incline (downwards) = mg sin(θ)

The horizontal force can be resolved into two components

Normal to incline = F sin(θ)

along incline (upwards) = F cos(θ)

Net force pushing mass up incline

= F cos(θ) - mg cos(θ)

= 22 cos(30) - 2.86*9.81 sin(30)

= 19.053 - 14.028

= 5.024 N up incline

Net acceleration up incline

= 5.024 / 2.86

= 1.757 m/s/s

initial velocity = 0

final velocity = v m/s

distance = d = 0.73 m

v^{2}- 0^{2}= 2ad

v^{2}

= 2*1.757*0.73

= 2.565

v=sqrt(2.565)

=1.60 m/s

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