Force pulling a box at 25° over 12 m with friction

[punjabi_monster]

I am having trouble with this physics question, i would appreciate it if some could show me how it can be solved.

A 1.50 * 10^2 N force is pulling a 50.0 kg box along a horizontal surface. The force acts at an angle of 25.0°. If this force acts through a displacement of 12.0 m, and the coefficient of friction in 0.250, what is the speed of the box, assuming it started from rest?

 

[Parth Dave]

How did you attempt to solve this problem? Where specifically are you having trouble?

 

[JasonRox]

Do not post the question everywhere!

 

[punjabi_monster]

this is how i attempted to solve the question:

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.

 

[Pyrrhus]

You seem to have forgotten

The force acts at an angle of 25.0°

this force has a vertical component.

 

[punjabi_monster]

how do i solve it by incoporating the vertical component?

 

[Pyrrhus]

1.50 * 10^2*sin(25) = Fy

 

[punjabi_monster]

i still do not understand.
after you find fy, then where do you incorporate taht in finding the velocity?

 

[Pyrrhus]

Sorry, i was helping someone else, Fy is pointing up as the normal force, so Fy+ n= mg

 

[punjabi_monster]

k tahnks i got the answer

 

[Pyrrhus]

k tahnks i got the answer

no problem, it's good to be of help.