A 170 g hockey puck sliding at 25 m/s slows to a speed of 10 m/s over a distance of 7.5 m. Determine the force of friction causing the puck to slow.
Answer (2)
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= 2Δ d v k 2 − v p 2 a= 2 ∗ 7 , 5 100 − 625 a= 15 − 525 a=-35 s m ²
F=0,17kg*(-35m/s²)=-5,95N
To find the force of friction acting on a hockey puck that slows from 25 m/s to 10 m/s over 7.5 m, we calculated the acceleration using kinematics and then applied Newton's second law. The force of friction is approximately -5.95 N, indicating it acts opposite to the direction of motion. The mass of the puck is 0.17 kg, and the acceleration calculated was -35 m/s².
;
