Hockey pucks - V t 1/2at 2 o ΔX = 0 ½(4m/s 2(5.5 s 2 ΔX...

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28). A hockey puck (mass = 1.5 kg) leaves the players stick (moving to the left) with a speed of 22 m/s and slides on the ice before coming to rest. The coefficient of friction between the puck and the ice is 0.4. What is the normal force on the puck? o F W1E = m*g o F W1E = 1.5kg * 10 m/s 2 15 N What is the friction force exerted on the puck due to the ice? o F W1E * μ = F μ due to ice o 15 N * 0.4 6 N What is the magnitude of the acceleration of the puck due to this friction force? o F= m*a o 6 N = 1.5 kg * a a= 4 m/s 2 How long will the puck slide after leaving the players stick? (what amount of time will it be sliding) o V = V 0 + a*t o 22 m/s = 0 + (4 m/s 2 ) * t t= 5.5 s How far will the puck slide after leaving the players stick? o ΔX =
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Unformatted text preview: V t + 1/2at 2 o ΔX = 0 + ½ (4m/s 2 )(5.5 s) 2 ΔX = 60.5 m A hockey puck (mass = 3.5 kg) leaves the players stick with a speed of 35 m/s and slides on the ice for 90 meters before coming to rest. What is the magnitude of the acceleration on the puck? o V 2 = V 2 + 2*a*ΔX 35 2 = 0 + 2*a* 9m o 6.8056 m/s 2 What is the friction force exerted on the puck due to the ice? o f 1S = ma 3.5 kg * 6.8056 m/s 2 o 23.8194 N What is the normal force on the puck? o F N1S = m*g 3.5 kg * 10 m/s 2 o 35 N What is the friction coeFcient between the puck and the ice? o Force exerted due to ice / normal force 23.8194 N/ 35 N o µ = 0.6806 (unitless)...
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