Question

A block of mass m1 = 5 kg on a frictionless horizontal table is connected to a block of mass m2 = 3 kg by means of a very light pulley P1 and a light fixed pulley P2 as shown in figure. If a1 and a2 are the accelerations of m1 and m2, respectively, (a) what is the relationship between these accelerations? Find (b) the tensions in the strings and (c) the accelerations a1 and a2. Take g = 10 ms-2
Only legitimate answers please

Answer 1

Answer:

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Answer 2

Given:

A block of mass m1 = 5 kg on a frictionless horizontal table is connected to a block of mass m2 = 3 kg by means of a very light pulley P1 . If a1 and a2 are the accelerations of m1 and m2, respectively.

To Find:

(a) what is the relationship between these accelerations?

(b) the tensions in the strings ?

(c) the accelerations a1 and a2.  [Take g = 10 m/s-2]

Solution:

it is given that - mass of block A, m1=5kg

mass of block B, m2=3kg

and a1 and a2 are the accelerations of mass m1 and m2.

therefore, according to question,

here, for the block of mass m1;

$T - f_{c}=m_{1}a_{1}\\T -\mu m_{1}g=m_{1}a_{1}\\T=\mu m_{1} g+m_{1}a_{1}$.         ....(1)

for the block of mass m2;

$m_{2}g - T=m_{2}a_{2}\\T=m_{2}g-m_{2}a_{2}$          ....(2)

from equation (1) and (2), we get

$\mu m_{1}g+m_{1}a_{1}=m_{2}g-m_{2}a_{2}\\\\m_{1}a_{1}=m_{2}g-\mu m_{1}g-m_{2}a_{2}\\\\a_{1}=\dfrac{m_{2}}{m_{1}} g - \mu g \dfrac{m_{1}}{m_{1}} - \dfrac{m_{2}}{m_{1}}a_{2}\\\\a_{1}=\dfrac{m_{2}}{m_{1}} g - \mu g - \dfrac{m_{2}}{m_{1}}a_{2}\\\\a_{1}=K-Ta_{2} \\\\where, \ K = \dfrac{m_{2}}{m_{1}}g-\mu g \ ,\ T=\dfrac{m_{2}}{m_{1}}$

where, K and T are constants.

Hence,

(a) relationship between these accelerations is

$a_{1}=K-Ta_{2} \ , where \ K \ and \ T \ are \ constants.$

(b) tensions in the string is given by

$T=\dfrac{m_{1}m_{2}}{m_{2}-\mu m_{1}}(\mu a_{2}+a_{1})$

(c)the acceleration a1 and a2 is;

$a_{1}= \dfrac{T}{m_{1}}- \mu g$

and

$a_{2}=g - \dfrac{T}{m_{2}}$