In an ice rink a skator is moving at $3 m s$ $^{- 1}$ and encounters a rough patch that reduces her speed by $45 %$ due to a friction force that is $25 %$ of her weight. Find the length of the rough patch.

1.56 m

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1.46 m

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1.36 m

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1.26 m

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Solution

The correct option is D $1.26 m$
$l = \frac{v_{i}^{2} - v_{f}^{2}}{2 μ g}$

$= \frac{3^{2} (1 - (.55)^{2})}{2 \times 0.25 \times 10}$

$= 1.8 [.7] = 1.26 m$

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In an ice rink a skator is moving at 3ms−1 and encounters a rough patch that reduces her speed by 45% due to a friction force that is 25% of her weight. Find the length of the rough patch.

The correct option is D 1.26ml=v2i−v2f2μg=32(1−(.55)2)2×0.25×10=1.8[.7]=1.26m

In an ice rink a skator is moving at $3 m s$ $^{- 1}$ and encounters a rough patch that reduces her speed by $45 %$ due to a friction force that is $25 %$ of her weight. Find the length of the rough patch.

The correct option is D $1.26 m$
$l = \frac{v_{i}^{2} - v_{f}^{2}}{2 μ g}$

$= \frac{3^{2} (1 - (.55)^{2})}{2 \times 0.25 \times 10}$

$= 1.8 [.7] = 1.26 m$