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Sagot :
Answer: [tex]5.64\ m/s^2[/tex]
Explanation:
Given
mass of sled m=5 kg
Elevation [tex]\theta =45^{\circ}[/tex]
[tex]\mu _k=0.185[/tex] (coefficient of kinetic friction)
Friction will oppose the motion of the sled while sled weight sin component helps it to move down the incline
[tex]W\sin\theta-F_r=ma\\where\\W=weight(mg)\\F_r=Friction(\mu_kmg\cos\theta)\\a=acceleration[/tex]
Substituting values
[tex]mg\sin \theta-\mu_kmgcos\theta=ma\\g\sin \theta-\mu_kg\cos \theta=a\\a=g(\sin 45^{\circ}-0.185\times \cos 45^{\circ})\\a=5.64\ m/s^2[/tex]
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