Work-mechanical energy: box sliding

[mamii92]

Homework Statement

In Figure 7-47a, a 1.7 N force is applied to a 5.6 kg block at a downward angle θ = 49° as the block moves rightward through 1.4 m across a frictionless floor. Find the speed vf of the block at the end of that distance if the block's initial velocity is (a) 0 and (b) 1.3 m/s to the right. (c) The situation in Figure 7-47b is similar in that the block is initially moving at 1.3 m/s to the right, but now the 1.7 N force is directed downward to the left. Find the speed vf of the block at the end of the 1.4 m distance.

[the figure is showing a box with the force directed 40 degrees below the horizontal to the right. kind of like a clock pointing to 4 oclock)

Homework Equations

w=F.d
w= mvf^2/2 -mvi^2/2
Vf^2 -Vi^2 = 2ad
F=ma

The Attempt at a Solution

I tried this question using both kinematics equations and work mechanical energy theorem and i am getting the same answer but they are both wrong.

First, for a) Vi=0 so i used the equation Vf^ = 2ad.
to find a, i used the equation a= F/m, where F would be Fsintheta.= 1.283006286
so a= 1.283006286 / 5.6 = 0.229108265 m/s^2

thus, Vf = sprt.(2 x 0.229108265 x 1.4)=0.800938913 m/

I also used w= f.d which gave 1.7962088 J
then equating w=mvf^2/2, and Vf is still 0.800938913 m/s?
someone please help me finding out where i went wrong.

 

[ehild]

To calculate either the work or the horizontal acceleration, you need the horizontal component of the force. How do you get it?

ehild

 

[mamii92]

OH okay. i think the mistake was that i was using the vertical component fsintheta rather than the horizontal Fcostheta