Chapter 7: Q.55 (page 180)

What is the acceleration of the $3.0 k g$ block in $F I G U R E C P 7.55$ across the frictionless table?
Hint: Think carefully about the acceleration constraint.

Short Answer

Expert verified

The acceleration of the $3.0 k g$ block is $2.8 m / s^{2}$

Step by step solution

Given information

mass of block $m_{1} = 3.0 k g$

mass of block $m_{2} = 1.0 k g$

All surfaces are frictionless.

Explanation

Consider the Forces of two masses given in the $f i g u r e C P 7.55$

For block mass $m_{1}$

$T_{1} = m_{1} a_{1}$

For block mass $m_{2}$

$m_{2} g - T_{2} = m_{2} a_{2}$

Considering the pulley is massless, $∴ T_{1} = T_{2}$

Let $a_{2} = a$ $∴ T = 2 m_{1} a$

$∵ a_{1} = 2 a$ acceleration constraint

On substituting the above equations on block mass $m_{2}$ equation, we get

$m_{2} g - 2 m_{1} a = m_{2} a (2 m_{1} + m_{2}) a = m_{2} g a = \frac{m_{2} g}{(2 m_{1} + m_{2})}$

Therefore, the acceleration for $m_{1}$ is given by,

localid="1649865147826" $a_{1} = \frac{2 m_{2} g}{(2 m_{1} + m_{2})} = \frac{2 (1.0) (9.81)}{(2 (3.0) + 1.0)} = 2.8 m / s^{2}$

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What is the acceleration of the $3.0 k g$ block in $F I G U R E C P 7.55$ across the frictionless table?
Hint: Think carefully about the acceleration constraint.

Short Answer

Step by step solution

Given information

Explanation

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What is the acceleration of the 3.0kgblock in FIGURECP7.55across the frictionless table?Hint: Think carefully about the acceleration constraint.

Short Answer

Step by step solution

Given information

Explanation

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What is the acceleration of the $3.0 k g$ block in $F I G U R E C P 7.55$ across the frictionless table?
Hint: Think carefully about the acceleration constraint.