A 1200-kg car rolling on a horizontal surface has a speed v= 65km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?
To find the spring stiffness constant of the spring, we can use the principle of conservation of mechanical energy. When the car strikes the spring, its kinetic energy is transferred to the spring potential energy.
First, let's convert the car's speed from km/h to m/s:
Given: v = 65 km/h
1 km/h = 0.2778 m/s(approximately)
65 km/h * 0.2778 m/s = 18.06 m/s
Next, let's calculate the initial kinetic energy (KE) of the car:
KE = ½ mv^2
Given: mass (m) = 1200 kg
v = 18.06 m/s
KE = ½ * 1200 kg * (18.06 m/s)^2
Now, we can calculate the spring potential energy (PE) using the displacement (d) and the spring constant (k):
PE = ½ kx^2
Given: displacement (d) = 2.2 m
PE = ½ * k * (2.2 m)^2
According to the conservation of mechanical energy, the initial kinetic energy is equal to the spring potential energy, so we equate these two equations:
½ * 1200 kg * (18.06 m/s)^2 = ½ * k * (2.2 m)^2
Now, we can solve for the spring constant (k):
k = (1200 kg * (18.06 m/s)^2) / (2 * (2.2 m)^2)
Evaluating this expression, we get the spring stiffness constant (k).
KEcar=PEspring
1/2 m v^2= 1/2 k x^2
solve for k.
8.1*10^4
PE=1/2kx
8.1x10^4