Respuesta :
Answer:
The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.
Explanation:
On a flat road, the springs just support the car's weight.
The weight of the car = force on the springs
And from Hooke's law, F = kx
On a flat road,
k x = mg
x(flat) = 1300 Ă— 9.8/k = 12740/k
Through the bottom of the circular dip, the springs not only balance the weight of the car, they also balance the force that keeps it in circular motion.
kx = (mv²/r) + mg
kx = (1300 × 27²/600) + (1300×9.8) = 14319.5N
x(circular dip) = 14319.5/k
If the value of the springs' spring-constant is available, we'd get straight values for This, but in their absence,
x(circular dip)/x(flat road) = 14319.5/12740 = 1.12
The springs compress by an extra 12% when driven through the bottom of a circular dip compared to being driven on a flat road.
