Show all work dont forget units The 500kg coaster enters the

Show all work don\'t forget units: The 500kg coaster enters the power lift with a speed of 5.24 m/s at a height of 22.5 meters. It is then lifted to a height of 124.5m in 25.0 seconds and exits the lift at point 2 with a speed of 8.26 m/s. The coaster rolls down the incline and enters a loop. After the loop the coaster is brought safely to rest by a large spring that compresses 30.0 meters.

(a) Find the acceleration of the coaster as it enters the loop in g\'s,(b) the maximum safe height of the loop and (c) the spring constant of the spring that stops the coaster.

2)-- 8.26 ms 124.5 m 5.24 mis 22.5 m 30m

Solution

a) Speed at loacation 2 is 8.26 m/s.

Applying energy conservation to find speed at bottom,

PE + KE = constant

mgh + m(8.26^2) /2 = 0 + m v^2 / 2

(2 x 9.8 x 124.5) + (8.26^2 ) = v^2

v = 50.08 m/s

suppose speed at top is vf.

then , N + mg = m vf^2 / R

N = m vf^2 / R - mg > = 0

vf^2 / R = g

vf^2 = Rg

and using energy to conservation to find R.

m v^2 /2 + 0 = mg(2R) + mvf^2 /2

vf^2 = v^2 - (4gR)

v^2 - (4gR) = Rg

50.08^2 = 5gR

R = 51.18 m

so maximum heigh. h = 2R = 102.37 m .........Ans(B)

a) a = v^2 / R   = 50.08^2 / 51.18 = 49 m/s^2

in g: a = 49/9.8 = 5g

Ans: 5   ....Ans(a)

c) Using energy conservation for spring system,

mv^2 /2 + 0 = 0 + kx^2 /2

500 x 50.08^2 = k (30^2)

k = 1393.34 N/m .....Ans