- #1

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- Thread starter punjabi_monster
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- #1

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- #2

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Fg=mg

=(50kg)(-9.81m/s2)

= -491 N

Fnet=Fn-Fg

Fn= 491 N

Ff=uFn

=(0.250)(491N)

=123 N

W=Ek

W=Fd

W=(123 N)(12.0 m)

W=1472 J

Ek=1/2mv2

V=squareroot 2Ek/m

V=squareroot 2(1472 J)/(50.0 kg)

V= 7.7 m/s

The actual answer is 2.53 m/s. Can you tell me what i am doing wrong, thanks.

- #3

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im going to call the 1.5*10^2 N Fp

okay. the normal force is the weight minus the vertical component of the force, becaues the force is at an angle of 25 degrees, right? so its weight-Fpsin25

use that to calculate friction, mu*normal

horizontal force is then Fpcos25-Ffriction

divide that by 50 kg to get acceleration, reduces to a kinematics problem

you get?

- #4

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thanks i got the answer

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