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(7%) Problem 14: A block with mass m1 = 9.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of ?k = 0.61. A second block with a mass m2 = 10.2 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs. 20% Part (a) Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, ?Fy, for block 2. 20% Part (b) Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, ?Fx for block 1. 20% Part (c) Write an expression for the magnitude of the acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (Assume the cable connecting the masses is ideal.) 20% Part (d) Write an expression using the variables provided for the magnitude of the tension force, T. 20% Part (e) What is the tension, T in Newtons?

          (7%) Problem 14: A block with mass m1 = 9.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of ?k = 0.61. A second block with a mass m2 = 10.2 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs.

20% Part (a) Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, ?Fy, for block 2.

20% Part (b) Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, ?Fx for block 1.

20% Part (c) Write an expression for the magnitude of the acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (Assume the cable connecting the masses is ideal.)

20% Part (d) Write an expression using the variables provided for the magnitude of the tension force, T.

20% Part (e) What is the tension, T in Newtons?
        
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(7%) Problem 14: A block with mass m1 = 9.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of ?k = 0.61. A second block with a mass m2 = 10.2 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs.

20% Part (a) Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, ?Fy, for block 2.

20% Part (b) Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, ?Fx for block 1.

20% Part (c) Write an expression for the magnitude of the acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (Assume the cable connecting the masses is ideal.)

20% Part (d) Write an expression using the variables provided for the magnitude of the tension force, T.

20% Part (e) What is the tension, T in Newtons?

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Problem 14: A block with mass m1 = 9.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of μk = 0.61. A second block with mass m2 = 10.2 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs. Part (a) Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, ΣFy, for block 2. Part (b) Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, ΣFx for block 1. Part (c) Write an expression for the magnitude of the acceleration of block 2, a2, in terms of the acceleration of block 1, a1. (Assume the cable connecting the masses is ideal.) Part (d) Write an expression using the variables provided for the magnitude of the tension force, T. Part (e) What is the tension, T in Newtons?
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Transcript

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00:01 Here we have to solve the following problem.
00:03 First, we have to use variables to represent expression for the sum of the forces in y direction for the block 2.
00:16 Let's do this.
00:19 Block 2 is experiencing tension up and gravity down and overall it results in acceleration down.
00:28 Therefore, this can be written as following m2g minus t equals to m2 now let's answer question b.
00:41 Here we have to write down an equation and actually you know to be more precise we should place positive sign before t and negative sign before m to g and before m to a in the first equation but therefore we can multiply it by minus 1 and overall the result will be the same now let's answer question b we have to write down an equation for block 1 so this block is experiencing tension t and there is a kinetic friction f k so that is means that this equation is t minus mu k m1g equals to m1a now let's answer a question c here we have to down an equation for the magnitude of acceleration of block a 2 and here to be more precise yeah we should label each acceleration first as a 2 and a 1 respectively but in question c we notice that the magnitude of these vectors are the same therefore a 1 equals to a 2 equals to 8 now we can in question c additionally we can solve these two we can solve these two equations with respect to a.
02:37 So here we have two equations we'll label them as equation one and equation two and here we will calculate sum of one and two.
02:49 So sum of one and two is following so that is m2g minus mu k m1...
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