A block of mass m slides with a speed of $v_{0} = 6 m/s$ along a track from one level to a heigher level as shown. The track is frictionless until the block reaches the higher level where co-efficient of friction is $0.6$ . The distance $d$ travelled by block on higher level before stopping is $(g = 10 {m/s}^{2})$

\frac{7}{6} m

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\frac{5}{6} m

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\frac{29}{6} m

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3 m

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Solution

The correct option is A $\frac{7}{6} m$
By work energy therorem, $W_{m g} + W_{f r i c t i o n} = K_{f} - K_{i}$
$- m g h - μ mgd = 0 - \frac{1}{2} m v_{0}^{2}$
$10 \times 1.1 + 0.6 \times 10 \times d = \frac{1}{2} \times 36$
$d = \frac{7}{6} m$

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A block of mass m slides with a speed of v0=6 m/s along a track from one level to a heigher level as shown. The track is frictionless until the block reaches the higher level where co-efficient of friction is 0.6. The distance d travelled by block on higher level before stopping is (g=10 m/s2)

The correct option is A 76 mBy work energy therorem, Wmg+Wfriction=Kf−Ki −mgh−μmgd=0−12mv20 10×1.1+0.6×10×d=12×36 d=76 m

The correct option is A $\frac{7}{6} m$
By work energy therorem, $W_{m g} + W_{f r i c t i o n} = K_{f} - K_{i}$
$- m g h - μ mgd = 0 - \frac{1}{2} m v_{0}^{2}$
$10 \times 1.1 + 0.6 \times 10 \times d = \frac{1}{2} \times 36$
$d = \frac{7}{6} m$