when loading a toy launcher, a spring is pulled back 15 cm. the spring is known to have a spring constant of 315 N/m. A small object of mass 4.46 g is place at the end of the spring. What acceleration will the object experience the instant the spring is released?
f = m a ... a = f / m
a = (315 N * .15) / .00446 kg
a is in m/s^2
Could you explain why you multiplied the constant by the .15m?
that is the compression of the spring
the more it's compressed, the greater the force
To find the acceleration of the object the instant the spring is released, we can use Hooke's Law and Newton's Second Law of Motion.
Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. It can be written as:
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
Given that the spring constant (k) is 315 N/m and the displacement (x) is 15 cm (which we should convert to meters), we can find the force exerted by the spring:
x = 15 cm = 0.15 m
F = -kx = -(315 N/m)(0.15 m) = -47.25 N
Now, according to Newton's Second Law, the force acting on an object is equal to its mass multiplied by its acceleration:
F = ma
Rearranging the equation, we can solve for acceleration (a):
a = F / m
Given that the mass (m) of the object is 4.46 g (which we should convert to kilograms), we can calculate the acceleration:
m = 4.46 g = 0.00446 kg
a = (-47.25 N) / (0.00446 kg)
a ≈ -10588.34 m/s²
Therefore, the object will experience an acceleration of approximately -10588.34 m/s² (negative sign indicating it is in the opposite direction of the displacement of the spring) the instant the spring is released.
Note: The negative sign in the acceleration indicates that the object is accelerating in the opposite direction of the displacement of the spring.