Finding the velocity of an object being pushed up an angular slope

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A box with a mass of 1 kg is pushed up a 30-degree incline with a force of 11 N, and the coefficient of kinetic friction is 0.5. The work done against gravity is calculated as -9.8 J, while the work done by the applied force is found to be 22 J. The frictional force is determined to be 4.9 N, leading to a total work done of approximately 3.713 J. The calculations for velocity using the kinetic energy theorem yield a result of 2.72 m/s, but there are concerns about the accuracy of the normal force calculation. It is emphasized that the normal force is not zero and should be resolved into components perpendicular to the slope for accurate results.
Murph
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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.
Please help!
 

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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?
 
Would it not just be the angle of 30 degrees which the slope is angled at?
 
Murph said:
Would it not just be the angle of 30 degrees which the slope is angled at?
Nope, what is the defintion of Θ in the equation for work done by a constant force?
 
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
 
Murph said:
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)
Looks better to me :approve:

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.
 
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?
 
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.
 
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