Get insightful responses to your questions quickly and easily on IDNLearn.com. Ask your questions and receive comprehensive and trustworthy answers from our experienced community of professionals.
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]
We appreciate your contributions to this forum. Don't forget to check back for the latest answers. Keep asking, answering, and sharing useful information. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
