A car drives straight down toward the bottom of a valley and up the other side on a road whose bottom has a radius of curvature of 125mm . At the very bottom, the normal force on the driver is twice his weight.
To determine the speed of the car at the bottom of the valley, we can use the concept of conservation of energy.
At the top of the valley, the car has gravitational potential energy which is converted to kinetic energy as it moves down the incline. At the bottom of the valley, all the potential energy is converted to kinetic energy. Assuming no other forms of energy are involved (such as friction), we can equate the initial potential energy to the final kinetic energy.
The potential energy at the top of the valley can be given by:
PE_top = m * g * h
Where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height of the incline.
Since the car is traveling along a circular path with a radius of curvature, the height at the top of the valley is equal to the radius of curvature (125 mm).
PE_top = m * g * 0.125 m
The kinetic energy at the bottom of the valley is given by:
KE_bottom = (1/2) * m * v²
Where v is the speed of the car at the bottom.
Equating the initial potential energy to the final kinetic energy:
PE_top = KE_bottom
m * g * 0.125 = (1/2) * m * v²
Simplifying and solving for v:
v = √(2 * g * 0.125)
Plugging in the values:
v = √(2 * 9.8 * 0.125)
v = √(2.45)
v ≈ 1.56 m/s
Therefore, the speed of the car at the bottom of the valley is approximately 1.56 m/s.