Physics problem

1. Sep 15, 2004

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?

2. Sep 15, 2004

ahrkron

Staff Emeritus
What have you done so far?

3. Sep 15, 2004

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.

4. Sep 16, 2004

TenaliRaman

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

-- AI

5. Sep 16, 2004

cronxeh

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

( pull <-- object --> Ffriction)

Last edited: Sep 16, 2004