A block of mass 5 kg is kept over a rough horizontal surface. A time varying force acts on it along the horizontal given by F = 2t. The block starts slipping at t = 2.5 s and its acceleration at t = 3 s is $0.4 {m/s}^{2}$ . The summation of coefficients of static and kinetic friction is (take $g = 10 {ms}^{- 2}$ )

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Solution

$μ_{s} = \frac{2 t}{m g} = \frac{2 \times 2.5}{5 \times 10} = 0.1 (a t t = 2.5)$
(block just about to slide)
after 3 sec, $F - μ_{k} N = m a$
$2 \times 3 - μ_{k} . 5 \times 10 = \frac{5 \times 4}{10}$
$∴ μ_{k} = \frac{8}{100} = 0.08$

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Q. A block of mass 5 kg is kept over a rough horizontal surface. A time varying force acts on it along the horizontal given by F = 2t. The block starts slipping at t = 2.5 s and its acceleration at t = 3 s is

0.4 {m/s}^{2}

. The summation of coefficients of static and kinetic friction is (take

g = 10 {ms}^{- 2}

)

A block of mass 5 kg is lying on a rough horizontal surface. The coefficients of static and kinetic friction between the block and the surface respectively are 0.7 and 0.5. A horizontal force just sufficient to move the block is applied to it. If the force continues to act even after the block has started moving, what will be the acceleration of the block? Take $g = 10 m s^{- 2}$ .