Velocity of a mass being pulled across a level surface

[ttk3]

Homework Statement

A man pulls a crate of mass 67.0 kg across a level floor. He pulls with a force of 172.0 N at an angle of 28.0° above the horizontal. When the crate is moving, the frictional force between the floor and the crate has a magnitude of 124.0 N.
If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.70 m?

Homework Equations

W = Fsin(angle)
W = KE
KE = 1/2 m v^2

The Attempt at a Solution

I started by finding the net force so that I could use the work equation:

Fx = 172 cos(28) - 124 = 27.87 N
Fy = 172 sin(28) (Fn and W cancel) = 80.75 N
Fnet = sqrt(Fx^2 + fy^2) = 85.42 N

Work Equation

85.42(2.70)cos(28) = 203.6 J

KE Equation

203.6 J = .5 (67 kg) v^2

v= 2.47 m/s

 

[asd1249jf]

Careful, your frictional force acts in a opposing direction of your motion.. which is also angled.

And why does your $$F_n$$ and $$W$$ cancel? Weight comes straight down and your normal force is perpendicular to the surface.

Draw a free body diagram - it'll give you much better insight

 

[m2003]

Your attempt at a solution is correct. By using the work equation and the kinetic energy equation, you were able to find the final velocity of the crate after it has been pulled a distance of 2.70 m. This final velocity of 2.47 m/s is the result of the man's pulling force of 172.0 N at an angle of 28.0° above the horizontal, as well as the frictional force of 124.0 N between the floor and the crate. Additionally, your calculation of the net force was accurate, and your use of trigonometric functions was appropriate. Overall, your solution is well-supported and demonstrates a good understanding of the concepts of work, kinetic energy, and net force.