Question

10. A skateboarder goes down a ramp formed by the arc ofa circle dS 8m of radius 8 m. The total mass of the skateboarder and her board is 65 kg. She starts from rest at A, the top of the ramp. Find the speed with which she leaves the ramp at B, given that there is a constant resistance of 35 N 409 8m

Answer 1

Alright Dude, If that worked for you... dont forget to give THUMBS UP.(that will work for me!)

If I missed something feel free to leave a comment.
atleast before giving downvote.
and, Thanks for using Chegg- Smarter way to study.

Given, mass = 65 kg resistive force (F_R) = 35 N At point 'A'. total Energy = PE = m gh where, h = 8m {consider, potential energy w.r.t to bottom of path} So, E_A = m gh = (65 kg) (9.8) (8m) = 5096 J from A to B: length of arc = 130/360 times 2 pi R = 130/360 times 2 pi (8m) = 18.15 m So, work done again resistive force = (18.15 m) (35 N) = 635.3 N At point 'B': y = (8m - 8 cos 40 degree) = 1.87 m So, total energy at 'B' = KE + PE = 1/2 mv^2 + mg y Now, According to work - energy principle: Therefore E_A - W_R = E_B Therefore 5096 J = 635.3 J = 1/2 mv^2 + mg y Therefore 4460.7 = 1/2 mv^2 + (65) (9.8) (1.87) Therefore 1/2 mv^2 = 3269.51 Therefore v = squareroot 2 (3269.51)/65