Speed of a 50.0 kg Box Pulled by 1.50 * 10^2 N Force

[punjabi_monster]

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?

 

[ahrkron]

What have you done so far?

 

[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.

 

[TenaliRaman]

You have not accounted for the angle anywhere.
Resolve the applied force into its horizontal and vertical component.

-- AI

 

[cronxeh]

Ffriction=mu * Fn * cos (angle) -- this is the force due to friction, acting in opposite direction to pull

( pull <-- object --> Ffriction)