In the figure shown find the acceleration of 2 kg body and also take - s -0-5- - k -0-4- Take -g -10 m - s 2-B- 1 m - s 2C- 0-5 m - s 2D- 0 m - s 2

The correct option is D 0 m/s^2 nbsp;FBD of both the blocks nbsp; The maximum frictional force f_max=μ_s N=0.5×2× 10=10 N Tension in the string will be T= ...

In the figure...

Question

In the figure shown find the acceleration of $2 kg$ body and also take $μ_{s} = 0.5$ , $μ_{k} = 0.4$ $[$ Take $g = 10 {m/s}^{2}]$

1.5 {m/s}^{2}

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1 {m/s}^{2}

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0.5 {m/s}^{2}

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0 {m/s}^{2}

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Solution

The correct option is D $0 {m/s}^{2}$
FBD of both the blocks

The maximum frictional force $f_{m a x} = μ_{s} N = 0.5 \times 2 \times 10 = 10 N$
Tension in the string will be $T = 1 \times g = 10 N$
As $f_{m a x} = T$ hence the frictional force is sufficient enough to hold the system in equilibrium. So the sysem is in rest $(a = 0)$

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In the figure shown find the acceleration of 2 kg body and also take μs=0.5, μk=0.4[Take g=10 m/s2]

The correct option is D 0 m/s2 FBD of both the blocks The maximum frictional force fmax=μsN=0.5×2×10=10 N Tension in the string will be T=1×g=10 N As fmax=T hence the frictional force is sufficient enough to hold the system in equilibrium. So the sysem is in rest (a=0)

In the figure shown find the acceleration of $2 kg$ body and also take $μ_{s} = 0.5$ , $μ_{k} = 0.4$ $[$ Take $g = 10 {m/s}^{2}]$