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Standard XII

Physics

Calculating Friction Using Coeffcients

Two bodies of...

Question

Two bodies of mass $3 k g$ and $4 k g$ are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is $(g = 9.8 m / s^{2})$

4.9 m / s^{2}

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2.45 m / s^{2}

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1.4 m / s^{2}

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9.5 m / s^{2}

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Solution

The correct option is C $1.4 m / s^{2}$
$F_{n e t} = (4 g - 3 g)$

$g = m_{s y s} a_{s y s}$

$g = 7 K g \times a_{s y s}$

$a_{s y s} = \frac{10}{7} \Rightarrow 1.4 m / s^{2}$

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Similar questions

Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is $(g = 9.8 m / s^{2})$

Q. A light and smooth pulley is attached to the ceiling of a lift moving upward with acceleration

a_{0} = 2 {m/s}^{2}

as shown in the figure. Two blocks of equal masses

(m = 3 kg)

are attached to the two ends of a massless string passing over the pulley. Find the tension in the string. (Take

g = 10 {m/s}^{2}

Q. Bodies

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respectively are hanging as shown in figure. If the pulley is frictionless, find the common acceleration and the tensions in the two strings. Given:

g = 10 {m / s}^{2}

Q. Two bodies of masses

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and

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are attached to the ends of a string passing over a pulley. The

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mass is attached to the table top by another string. The magnitude of tension in this string

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is equal to (

g = 10 m / s^{2}

)

Q. The blocks of masses

m_{1} = 4 k g, m_{2} = 2 k g, m_{3} = 4 k g

are connected with ideal strings passing over a smooth massless pulley as shown in the figure. The acceleration of blocks will be ?

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Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g=9.8m/s2)

The correct option is C 1.4 m/s2Fnet=(4g−3g)g=msysasysg=7Kg×asysasys=107⇒1.4m/s2

Two bodies of mass $3 k g$ and $4 k g$ are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is $(g = 9.8 m / s^{2})$

The correct option is C $1.4 m / s^{2}$
$F_{n e t} = (4 g - 3 g)$

$g = m_{s y s} a_{s y s}$

$g = 7 K g \times a_{s y s}$

$a_{s y s} = \frac{10}{7} \Rightarrow 1.4 m / s^{2}$