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carljohnston2
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Hi there, I need help with understanding this guy's solution thanks!
The kinetic friction force will be up the slide to oppose the motion.
We choose the positive direction in the direction of the acceleration.
From the force diagram for the child, we have ?F = ma:
x-component: mg sin [tex]\theta[/tex] – Ffr = ma;
y-component: FN – mg cos [tex]\theta[/tex] = 0.
When we combine these, we get
a = g sin [tex]\theta[/tex] – [tex]\mu[/tex]kg cos [tex]\theta[/tex] = g(sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex]).
We can use this for the frictionless slide if we set [tex]theta[/tex]k = 0.
For the motion of the child, we have
v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide.
If we form the ratio for the two slides, we get
(vfriction/vnone)2 = afriction/anone = (sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex])/sin [tex]\theta[/tex];
(!)2 = (sin 28° – [tex]\mu[/tex]k cos 28°)/sin 28°, which gives [tex]\mu[/tex]k = 0.40.
The kinetic friction force will be up the slide to oppose the motion.
We choose the positive direction in the direction of the acceleration.
From the force diagram for the child, we have ?F = ma:
x-component: mg sin [tex]\theta[/tex] – Ffr = ma;
y-component: FN – mg cos [tex]\theta[/tex] = 0.
When we combine these, we get
a = g sin [tex]\theta[/tex] – [tex]\mu[/tex]kg cos [tex]\theta[/tex] = g(sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex]).
We can use this for the frictionless slide if we set [tex]theta[/tex]k = 0.
For the motion of the child, we have
v2 = v02 + 2a(x – x0) = 0 + 2ad, where d is the distance along the slide.
If we form the ratio for the two slides, we get
(vfriction/vnone)2 = afriction/anone = (sin [tex]\theta[/tex] – [tex]\mu[/tex]k cos [tex]\theta[/tex])/sin [tex]\theta[/tex];
(!)2 = (sin 28° – [tex]\mu[/tex]k cos 28°)/sin 28°, which gives [tex]\mu[/tex]k = 0.40.
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