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A sled and rider, gliding over horizontal, frictionless ice at 4.3 m/s , have a combined mass of 74 kg . the sled then slides over a rough spot in the ice, slowing down to 2.9 m/s . part a what impulse was delivered to the sled by the friction force from the rough spot?

Respuesta :

skyluke89
The impulse of a force is defined as
[tex]I=F \Delta t[/tex]
where F is the intensity of the force and [tex]\Delta t[/tex] the time of application of this force.

We can rewrite the previous relationship by using Newton's second law:
[tex]F= ma[/tex]
substituting, the equation for the impulse becomes
[tex]I = m a \Delta t[/tex]

But the acceleration is the variation of the velocity in the time interval:
[tex]a= \frac{\Delta v}{\Delta t} [/tex]
so we can rewrite I as
[tex]I = m \frac{\Delta v}{\Delta t} \Delta t = m \Delta v [/tex]

the combined mass of sled and rider is m=74 kg, while the variation of velocity is 
[tex]\Delta v = 2.9 m/s - 4.3 m/s = -1.4 m/s[/tex]
and so we can calculate the impulse of the friction force:
[tex]I= (74 kg)(-1.4 m/s)=-103.6 kg m/s[/tex]
where the negative sign means the friction force acts against the motion, to decelerate the system.

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