A catapult used to hold a stone of mass 500g is
extended by 20cm with an applied force f. If the
force applied is 200N, what is the spring constant of the catapult?
The formula for Hooke's Law is F = kx, where F is the force, k is the spring constant, and x is the displacement.
First, we need to convert the mass of the stone from grams to kilograms:
mass (kg) = mass (g) / 1000 = 500g / 1000 = 0.5kg
Next, we need to convert the extension length from centimeters to meters:
extension (m) = extension (cm) / 100 = 20cm / 100 = 0.2m
We can plug in the values we have into Hooke's Law formula and solve for the spring constant:
200N = k * 0.2m
Dividing both sides of the equation by 0.2m:
k = 200N / 0.2m = 1000 N/m
Therefore, the spring constant of the catapult is 1000 N/m.
catapult follows Hooke's Law, which states that the force required to extend or compress a spring is directly proportional to the displacement, we can calculate the force applied using the equation:
F = k * x
Where:
F = Force applied (N)
k = Spring constant (N/m)
x = Displacement (m)
Given that the stone has a mass of 500g (0.5kg) and the catapult is extended by 20cm (0.2m), we can calculate:
F = k * x
F = 0.5kg * 9.8m/s^2 (acceleration due to gravity) (Weight of the stone)
F = 4.9N
Now, we need to calculate the spring constant (k). To do this, we can rearrange the equation:
k = F / x
k = 4.9N / 0.2m
k = 24.5 N/m
Therefore, the applied force (F) required to extend the catapult by 20cm is 4.9N, and the spring constant (k) is 24.5 N/m.