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If an object is moving at #24 m/s# over a surface with a kinetic friction coefficient of #u_k=6 /g#, how far will the object continue to move?

Answer 1

Layla Ahmed

#"48 m"#

Kind friction (#f#) acting on object #= μ_kmg#

Retardation (a) of object #= f/m = μ_k g"#

Object will move to a distance #S#

#S = u^2 / "2a" = u^2 / (2mu_kg) = ("24 m/s")^2 / (2 × 6/cancel"g" xx cancel"g") = "48 m"#

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Answer 2

Adrian Ayala

To find the distance the object will continue to move, we can use the formula:

[ d = \frac{v^2}{2 \cdot \mu_k \cdot g} ]

Where:

( d ) is the distance traveled,
( v ) is the initial velocity of the object,
( \mu_k ) is the coefficient of kinetic friction,
( g ) is the acceleration due to gravity (approximately ( 9.8 , m/s^2 )).

Plugging in the given values:

[ d = \frac{(24 , m/s)^2}{2 \cdot 6/g \cdot 9.8 , m/s^2} ]

[ d = \frac{576 , m^2/s^2}{117.6/g} ]

[ d = \frac{576 , m^2/s^2 \cdot g}{117.6} ]

[ d \approx \frac{576 \cdot 9.8 , m}{117.6} ]

[ d \approx \frac{5644.8 , m}{117.6} ]

[ d \approx 47.99 , m ]

So, the object will continue to move approximately 47.99 meters.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.