∆KE if angle between force and displacement varies

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patrykh18
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Homework Statement



A string is attached to a block that can slide along a frictionless horizontal surface aligned along the x axis. The left end of the string is pulled over a pulley of negligible mass and friction at a height of 1.3m so that the block slides from x1=3m to x2=1m. The tension in the string remains constant (25N). Calculate the ∆KE.

Homework Equations



W=F•x=|F||x|cosA where A is the angle
KE=0.5mv^2
∆KE= KE final- KE initial

The Attempt at a Solution



In my first attempt I got an answer of 30.49 J. What I did is I I said that ma=-25cosA so a=-(25cosA)/m
I integrated with respect to t. Since the velocity at x=3m is not stated I let it equal v0. So the integration constant was v0. I integrated v with respect to time. My integration constant was 3 as that was the original displacement. Next I found the time at which x=1m in terms of m (I knew the angle at that time using basic trigonometry). I took my value of time and plugged it back into equation for velocity.

After doing it I realized that obviously my answer is wrong as I treated cosA like a constant as I was intergrating. My problem is that i can't figure out the expression for A as a function of time.

I have attached below a diagram of the system. Any help will be much appreciated.

1507988661517-1730420136.jpg
 
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Orodruin said:
Hint: You do not need to compute the velocity or how the position depends on time. In fact, you do not need to involve time at all in this problem. What is the relation between the work done on the block and the difference in its kinetic energy?

Well they are equal to each other. I am just not sure how to calculate the work done
 
Orodruin said:
What is the definition of work done?

Its the transfer of energy to or from an object due to the action of a force.
 
Orodruin said:
And how is it expressed in terms of the force and the displacement of the object?
It's the dot product of the two
 
It is displaced in the negative x direction and the force in that direction is negative 25cosA
 
Orodruin said:
So how can you express the angle in terms of the position? Can you give an expression for the integral you need to compute in order to find the work?
I suppose the angle can be expressed as Tan inverse of 1.3/x
 
Or it can also be expressed as cos inverse of x/√x^2 +1.3^2