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saplinglearning.com ackboard Learn Math 26a Papercut LogLte university Gmail Apple YouTube News ESPN Google www.saplinglearning.com/ibis/css/Accessibilityp. www.saplinglearning.com/ibis/css/Accessiblity www.saplinglearr Jump to... (? Ab ity, Sacramento- PHYS 5A-Fall17-0SBORNE Activities and Due Dates >HW: Applications of Forces 6/2017 11:00 PM 65.8/100 Grade Print @ Calculator- Periodic Table on 10 of 11 Sapling Learning Map shown below. The coefficient of kinetic friction between the box and the ramp is 0.145, and the rope pulling the box is parallel to the ramp· If the box accelerates up the ramp at a rate of 2.29 ms, what must the tension Fr in the rope be? Use g-9.81 m/s2 for the acceleration due to gravity. Number a=2.29 m/s m=6.65 kg -0.145 = 35.0° .0 Previous ) Check Answer 0 Next -a Exit Hint aboutus | careers privacy policy terms of use contact yExplanation / Answer
Forces in vertical direction,
N - mgcos theta = 0
N = mg cos theta
Forces in horizontal direction,
Ft - mgsin theta - Ff = ma
where, Ff = uN
Ft = ma + mgsin theta + uN
= ma + mgsin theta + umg cos theta
= ma + mg(sin theta + ucos theta)
= 6.65*2.29 + 6.65*9.8(sin 35 degree + 0.145*cos 35 degree)
= 60.35 N answer
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