Block is initially at rest- it is pulled by 2 t - - s -0-4 - k -0-38- Force 2 t is gradually increasing- then at t -8 sec -4 kg - F -2 t right w is weight- fr - frictional force A- F - N - fr - wB- N - fr - w - FC- F - N - f r - wD- body is accelerated

The correct option is A F⃗+N⃗+f⃗r⃗= w⃗ nbsp;f_net=μsN = 0.4 ×4×10= 16 N = Fapplied(at t=8 s) . So, the body will not move. since the forces are in equilibri ...

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Standard XII

Physics

The Complete Equilibrium

Block is init...

Question

Block is initially at rest, it is pulled by 2t, $μ_{s} = 0.4$ $μ_{k} = 0.38$ . Force 2t is gradually increasing, then at t=8 sec:

( $\to w$ is weight, $\to f r$ =frictional force)

\to F + \to N + \to f r = - \to w

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\to F + \to N + \to f r = \to w

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\to N + \to f r + \to w = \to F

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body is accelerated

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Solution

The correct option is A $\to F + \to N + \to f r = - \to w$

$f_{n e t} =$ $μ_{s}$ N = 0.4 $\times$ 4 $\times$ 10= 16 N = $F_{a p p l i e d}$ (at t=8 s) . So, the body will not move. since the forces are in equilibrium , using parallelogram vector addition law, $\to F + \to N + \to f r = - \to w$

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Block is initially at rest, it is pulled by 2t, μs=0.4 μk=0.38. Force 2t is gradually increasing, then at t=8 sec: (→w is weight, →fr =frictional force)

The correct option is A →F+→N+→fr=−→w fnet=μsN = 0.4 ×4×10= 16 N = Fapplied(at t=8 s) . So, the body will not move. since the forces are in equilibrium , using parallelogram vector addition law, →F+→N+→fr=−→w

Block is initially at rest, it is pulled by 2t, $μ_{s} = 0.4$ $μ_{k} = 0.38$ . Force 2t is gradually increasing, then at t=8 sec:

( $\to w$ is weight, $\to f r$ =frictional force)

The correct option is A $\to F + \to N + \to f r = - \to w$

$f_{n e t} =$ $μ_{s}$ N = 0.4 $\times$ 4 $\times$ 10= 16 N = $F_{a p p l i e d}$ (at t=8 s) . So, the body will not move. since the forces are in equilibrium , using parallelogram vector addition law, $\to F + \to N + \to f r = - \to w$