# Homework Help: Finding the velocity of an object being pushed up an angular slope

1. Nov 4, 2008

### Murph

1. A box mass m=1 kg is pushed up an incline of an angle theta=30 degrees that has a coefficient of kinetic friction u_k=.5. Find the velocity of the object after it pushed for d=2m by a force of magnitude F=11N directed upward alone the incline.

W_g=m*g*d*cos(90+theta)
W_constant force=F*d*cos(theta)
F_f=m*g*u_k
F_n=m*g
W(which is the sum of the work done on the box)=1/2*m*v^2

First off I solved for the W_g, which is just straight substitution W_g=1*9.8*2*cos(120)=-9.8. After that I solved for the constant force W_constant force= 11[N]*2*cos(30)=19.05. Now this is the part where I try to determine the work that friction affects the box which is F_f=1*9.8*.5=4.9 and I am not for sure if I need to solve for the work done by friction W_f=4.9*2*cos(30) or not and i know the normal force is 0 since F_n is perpendicular to the displacement .... I know the answer is 3.2 [M/s] but I am off by a bit every time.

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2. Nov 4, 2008

### Hootenanny

Staff Emeritus
Let's consider your expression for the work done by the constant force for a moment. You have

W = Fd.cos(Θ)

What is the definition of the angle Θ?

3. Nov 4, 2008

### Murph

Would it not just be the angle of 30 degrees which the slope is angled at?

4. Nov 4, 2008

### Hootenanny

Staff Emeritus
Nope, what is the defintion of Θ in the equation for work done by a constant force?

5. Nov 4, 2008

### Murph

Oh yea... I have been poking in numbers for a while i did the angle of that to be zero as well so then my work done for the constant is W_c=11*2*cos(0degrees).. or am I still wrong... haha..which is 22

6. Nov 4, 2008

### Hootenanny

Staff Emeritus
Looks better to me

Θ is the angle between the applied force and the displacement, in this case the force and displacement are parallel and hence the angle is zero.

7. Nov 4, 2008

### Murph

ok so I have W_g=-9.8 my W_c=22 and my W_friction=-8.487, so then I add those sums to get my total work done which is W=3.713 then I plug that into the Kinetic energy theorem and have the v=sqrt(2*W/m)....sqrt(2*3.713/1)=2.72....and the answer is 3.2[M/s]....am I still making errors?

8. Nov 4, 2008

### Hootenanny

Staff Emeritus
You may want to recheck your calculation of the normal force. The normal is not zero as you stated in your opening post, and it is not simply the weight of the object as you have in your calculations.

Start by resolving the forces acting on the block into the components perpendicular to the slope.