# Physics question

I am having trouble with this physics question, i would appriciate 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?

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

JasonRox
Homework Helper
Gold Member
Do not post the question everywhere!

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
Homework Helper
You seem to have forgotten
The force acts at an angle of 25.0°
this force has a vertical component.

how do i solve it by incoporating the vertical component?

Pyrrhus
Homework Helper
1.50 * 10^2*sin(25) = Fy

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

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

k tahnks i got the answer

Pyrrhus
Homework Helper
punjabi_monster said:
k tahnks i got the answer

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