The carton shown in Fig. 4-55 lies on a plane tilted at an angle = 21.5° to the horizontal, with µk = 0.17.

(a) Determine the acceleration of the carton as it slides down the plane.

(b) If the carton starts from rest 8.80 m up the plane from its base, what will be the carton's speed when it reaches the bottom of the incline?

The carton shown in Fig. 4-55 lies on a plane tilted at an angle = 21.5° to the horizontal, with μk = 0.17.
OK , I'm late with an answer, I think i can improve on Ramble On's excellent answer by appealing to the concept of converting Potential to Kinetic energy on the 8.8 m trip down the incline. Choose m=1. PE=g*h=9.81*8.80*sin(21.5)=31.64J.



Nf=9.8*cos21.5=9.12

Ff=.17*9.12=1.55 N

Tf=9.8*sin21.5 Resultant F = Tf-Ff=2.04N/1=2.04m/s^2



Now converting all the Potential energy to kinetic energy,

.5*1*v^2=31.64

v^2=63.28

v=7.95 m/s