What is the speed of the box

  • #1
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?
 

Answers and Replies

  • #2
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.
 
  • #3
138
0
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?
 
  • #4
thanks i got the answer
 

Related Threads on What is the speed of the box

Replies
5
Views
1K
  • Last Post
Replies
1
Views
990
  • Last Post
Replies
2
Views
762
  • Last Post
Replies
23
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
6K
Replies
3
Views
1K
  • Last Post
Replies
15
Views
2K
Replies
1
Views
1K
Replies
4
Views
5K
Top