Case A:
A.75 kg 65 N/m 1.2 m
m = weight of the car = 0.75 kg
k = spring's stiffness = 65 N/m
h = elevation of the hill = 1.2 m
x = spring's compression = 0.25 m
Applying the principle of energy conservation from the Top of the hill to the Bottom of the hill
Energy at the Top of the hill equals Energy at the Bottom of the hill
spring energy + gravitational 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 = weight of the car = 0.60 kg
k = spring's stiffness = 35 N/m
h = height of the hill = 0.9 m
x = spring's compression = 0.25 m
Using energy conservation from the Top of the hill to the Bottom of the hill
Top hill energy = Bottom hill energy
spring energy + gravitational 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 = weight of the car = 0.55 kg
k = spring's stiffness = 40 N/m
h = height of the hill = 1.1 m
x = spring's compression = 0.25 m
Using conservation of energy from the Top of the hill to the Bottom of the hill
Energy at the Top of the hill = Energy at the Bottom of the hill
spring energy + gravitational 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 = weight of the car = 0.84 kg
k = spring's stiffness = 32 N/m
h = height of the hill = 0.95 m
x = spring's compression = 0.25 m
Using energy conservation from the Top of the hill to the Bottom of the hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + gravitational 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
thus, the closest result is from case C at 5.1 m/s
