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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.  

Verified 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=\(\frac{ vk^{2}- vp^{2} }{2Δd}\)
a=\(\frac{100-625}{2*7,5}\)
a=\(\frac{-525}{15}\)
a=-35\(\frac{m}{s}\)²
F=0,17kg*(-35m/s²)=-5,95N

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