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If an object is moving at #18 m/s# over a surface with a kinetic friction coefficient of #u_k=12 /g#, how far will the object continue to move?

Answer 1

Leo Abeyta

#x=13.5 " meters"#

#"1- Find the friction force :"# #F_f=u_k*m*g" "u_k=12/g#

#F_f=12/cancel(g)*m*cancel(g)#

#F_f=12m#

#"2- Find deceleration using Newton's law :"#

#F_f=m*a#

#12cancel(m)=cancel(m)*a#

#a=12 m/s^2#

#"3- Find distance using the formula :"#

#v_f^2=v_i^2-2*a*x#

#"where :"#

#v_i:"initial velocity"# #v_f:"final velocity=0"# #a:"deceleration"# #x:"distance"#

#0=18^2-2*12*x#

#324=24*x#

#x=324/24#

#x=13.5 " meters"#

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Answer 2

Leo Abeyta

To find the distance the object will continue to move, you can use the equation:

(d = \frac{v^2}{2 \mu_k g})

Where:

(d) is the distance
(v) is the initial velocity (18 m/s in this case)
(\mu_k) is the kinetic friction coefficient (0.12 in this case)
(g) is the acceleration due to gravity (approximately 9.8 m/s²)

Plugging in the values:

(d = \frac{(18 , \text{m/s})^2}{2 \times 0.12 \times 9.8 , \text{m/s}^2})

(d ≈ \frac{324}{23.52})

(d ≈ 13.78 , \text{meters})

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Answer from HIX Tutor

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