If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{12}{g}$ $u_k=12 /g$ , how far will the object continue to move?

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{12}{g}$ $u_k=12 /g$ , how far will the object continue to move?

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Answer 1 · 2015-12-13T19:48:53

Its a simple application of kinematics.

Explanation:

Normal force in this scenario is $N = m g$ $N= mg$
Frictional force $f = μ m g = u_{k} m g$ $f=\mu mg=u_kmg$
$u_{k} = \frac{12}{g}$ $u_k=\frac{12}{g}$ , therefore $f = 12 m$ $f=12m$ , where the $g$ $g$ s get cancelled.
Acceleration $a = 12 \frac{m}{s^{2}}$ $a=12m/s^2$
Therefore using the kinematic formula
$v^{2} - u^{2} = 2 a s$ $v^2-u^2=2as$ , where $s$ $s$ is the distance traveled, $u$ $u$ , $v$ $v$ are the initial and final velocities and $a$ $a$ acceleration.
$u = 4 \frac{m}{s}$ $u=4 m/s$
$v = 0$ $v=0$
$a = 12 \frac{m}{s^{2}}$ $a=12m/s^2$
Therefore, $s = \frac{u^{2}}{2 a}$ $s=u^2/{2a}$
substituting values given we get $s = \frac{4 \times 4}{2 \times 12} = \frac{16}{24} = \frac{2}{3} = 0.667 m$ $s={4\times4}/{2\times12}=16/24=2/3=0.667m$

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{12}{g}$ $u_k=12 /g$ , how far will the object continue to move?

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Explanation:

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If an object is moving at 4 m/s4ms4 m/s over a surface with a kinetic friction coefficient of u_k=12 /guk=12gu_k=12 /g, how far will the object continue to move?

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If an object is moving at $4 \frac{m}{s}$ $4 m/s$ over a surface with a kinetic friction coefficient of $u_{k} = \frac{12}{g}$ $u_k=12 /g$ , how far will the object continue to move?