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Byju's Answer

Standard XII

Physics

Tension in a String

The force act...

Question

The force acting on the block is given by $F = 5 - 2 t$ , the frictional force on the block after time $t = 2 s e c$ will be: $(μ = 0.2)$

2 N

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3 N

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1 N

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zero

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Solution

The correct option is A
2 N
$f_{m a x} = 10 \times 0.2 = 2 N$
Initial force = 5 N > 2 N
therefore block will move with acceleration
$a = \frac{5 - 2 t - f_{m a x}}{1} = 5 - 2 t - 2$
$\Rightarrow \frac{d v}{d t} = 3 - 2 t$
$\Rightarrow v = 3 t - t^{2}$
$\Rightarrow t = 0, 3 s e c, v = 0$
At t = 2 sec block is moving,
therefore $f_{m a x}$ will act i.e., frictional force acting = 2 N.

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The force acting on the block is given by F=5−2t, the frictional force on the block after time t=2 sec will be: (μ=0.2)

The correct option is A 2 Nfmax=10×0.2=2 N Initial force = 5 N > 2 N therefore block will move with acceleration a=5−2t−fmax1=5−2t−2 ⇒ dvdt=3−2t ⇒ v=3t−t2 ⇒ t=0,3 sec,v=0 At t = 2 sec block is moving, therefore fmax will act i.e., frictional force acting = 2 N.

The force acting on the block is given by $F = 5 - 2 t$ , the frictional force on the block after time $t = 2 s e c$ will be: $(μ = 0.2)$