Question
A catapult used to hold a stone of mass 500g is
extended by 20cm with an applied force f. If the
extended by 20cm with an applied force f. If the
Answers
GPT 3.5
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.
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.