A
170 g hockey puck sliding at 25 m/s slows to a speed of 10 m/s in a distance of
7.5 m.  Determine the force of friction
causing the puck to slow.  



Answer :

luana
From 2nd Newton Law:
F=ma
m=170g=0,17kg
Acceleration using kinematic equations:
vk²=vp²+2aΔd
vp-initial velocity=25m/s
vk-ending velocity=10m/s
Δd-distance
a=[tex] \frac{ vk^{2}- vp^{2} }{2Δd} [/tex]
a=[tex] \frac{100-625}{2*7,5} [/tex]
a=[tex] \frac{-525}{15} [/tex]
a=-35[tex] \frac{m}{s} [/tex]²
F=0,17kg*(-35m/s²)=-5,95N

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