∆KE if angle between force and displacement varies

In summary: I don't know how to do the integral. Can you help me?In summary, the homework statement is saying that a block is attached to a string and is pulled over a pulley. The tension in the string remains constant and the block slides from x1=3m to x2=1m. The relation between the work done on the block and the difference in its kinetic energy is not explained in the statement.
  • #1
patrykh18
32
2

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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  • #2
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?
 
  • #3
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
 
  • #4
What is the definition of work done?
 
  • #5
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.
 
  • #6
patrykh18 said:
Its the transfer of energy to or from an object due to the action of a force.
And how is it expressed in terms of the force and the displacement of the object?
 
  • #7
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
 
  • #8
So in this case, what is the force on the block at a given position and how is it being displaced?
 
  • #9
It is displaced in the negative x direction and the force in that direction is negative 25cosA
 
  • #10
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?
 
  • #11
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
 
  • #12
Or it can also be expressed as cos inverse of x/√x^2 +1.3^2
 

1. What is the relationship between ∆KE and the angle between force and displacement?

The relationship between ∆KE and the angle between force and displacement is described by the work-energy theorem. This theorem states that the change in kinetic energy (∆KE) of an object is equal to the work done on the object by external forces. In this case, the angle between force and displacement affects the amount of work done, which in turn affects the change in kinetic energy.

2. How does the angle between force and displacement affect the change in kinetic energy?

The angle between force and displacement affects the change in kinetic energy by changing the amount of work done on the object. When the force and displacement are parallel, the work done is at its maximum and the change in kinetic energy is also at its maximum. As the angle between force and displacement increases, the work done decreases and so does the change in kinetic energy.

3. Can the angle between force and displacement ever be negative?

No, the angle between force and displacement cannot be negative. This is because both force and displacement are vector quantities, meaning they have a magnitude and direction. The angle between two vectors is always positive and is measured from the direction of the first vector to the direction of the second vector.

4. How does the angle between force and displacement affect the direction of the change in kinetic energy?

The angle between force and displacement does not affect the direction of the change in kinetic energy. The direction of the change in kinetic energy is determined by the direction of the net force acting on the object. The angle between force and displacement only affects the magnitude of the change in kinetic energy.

5. Is there an optimal angle between force and displacement for maximizing the change in kinetic energy?

Yes, there is an optimal angle between force and displacement for maximizing the change in kinetic energy. This angle is 0 degrees, meaning the force and displacement are parallel. In this case, the work done on the object is at its maximum, resulting in the maximum change in kinetic energy. As the angle increases, the work done and change in kinetic energy decrease, until the angle reaches 90 degrees where there is no work done and therefore no change in kinetic energy.

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