Case A :
A .75 kg 65 N/m 1.2 m
m = mass of car = 0.75 kg
k = spring constant of the spring = 65 N/m
h = height of the hill = 1.2 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²
v = 5.4 m/s
Case B :
B .60 kg 35 N/m .9 m
m = mass of car = 0.60 kg
k = spring constant of the spring = 35 N/m
h = height of the hill = 0.9 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²
v = 4.6 m/s
Case C :
C .55 kg 40 N/m 1.1 m
m = mass of car = 0.55 kg
k = spring constant of the spring = 40 N/m
h = height of the hill = 1.1 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²
v = 5.1 m/s
Case D :
D .84 kg 32 N/m .95 m
m = mass of car = 0.84 kg
k = spring constant of the spring = 32 N/m
h = height of the hill = 0.95 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²
v = 4.6 m/s
hence closest is in case C at 5.1 m/s
