# What is the speed of the box

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?

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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.

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?
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?