A 120-kg box is loaded into the trunk of a car.
If the height of the car's bumper decreases by 20 cm , what is the force constant of its rear suspension?
looking for N/m
5900
Divide the box weight (M*g) by the deflection in meters (not cm)
With two significant figures, that is correct, in N/m.
To find the force constant of the car's rear suspension, we need to use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position (in this case, the change in height of the car's bumper).
Hooke's Law is given by the equation F = k * x, where F is the force exerted by the spring, k is the force constant (stiffness) of the spring, and x is the displacement.
In this scenario, the weight of the box loaded into the trunk acts as the force exerted by the spring in the car's rear suspension.
The weight of an object is given by the equation W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
Given that the mass of the box is 120 kg, we can calculate its weight by multiplying the mass by the acceleration due to gravity (approximately 9.8 m/s²):
W = 120 kg * 9.8 m/s²
W = 1176 N
Since we are looking for the force constant in N/m, we can rearrange Hooke's Law as k = F / x:
k = W / x
In this case, x is the change in height of the car's bumper, which is given as 20 cm. However, we need to convert it to meters:
x = 20 cm = 0.2 m
Now we can calculate the force constant:
k = 1176 N / 0.2 m
k ≈ 5880 N/m
Therefore, the force constant of the car's rear suspension is approximately 5880 N/m.