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dexar [7]
4 years ago
10

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

ock is pulled with a force of 100 N. What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?
Physics
1 answer:
o-na [289]4 years ago
8 0

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

F=ma

100=100m\times a

ma=1\ N

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

F_{100-99}=ma

F_{100-99}=1

We need to calculate the total number of masses attached to the string

Using formula for mass

m'=(100-50)m

m'=50m

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

F_{50}=m'a

Put the value into the formula

F_{50}=50m\times a

F_{50}=50\times1

F_{50}=50\ N

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

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3 years ago
A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,
Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

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Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

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\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )

\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times  \left ( v^{2}-0^{2} \right )

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