Answer:
a) 21.6 KN/m
b)v = 18.56 m/s
Explanation:
For this problem, the spring should be strong enough to get the car just over the top so:
mgh = (1/2)kx^2
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
a) K increased by Increase by 10%, so:
k = 19600 * 1.10 = 21560 N/m = 21.6KN/m---->specified spring constant
b) Now, Maximum speed of a 350 kg car will be at the bottom of the roller coaster:
therefore, (1/2)mv^2 + mg(Y_1-Y_2) = (1/2)kx^2
(0.5)*(350)*(v^2) + (350)(9.8)(10-15) =(0.5)*(21560)*(2.0^2)
v^2 = 344.4
v = 18.56 m/s