A pinball machine uses a spring that is compressed 4.0 cm to launch a ball. If the spring constant is 13 N/m, what is the force on the ball at the moment the spring is released?
I don't understand this I guess...
Does the ball leave the spring at the equalbarium position four cm from compression or does it travel past equalbarium and then the ball leaves? Also do I need to know the mass which I know how to do?
Could you please show me how to do this problem and the formulas to use... Thanks for the help
At the moment of release, the full force of the compressed spring is on the ball.
F = k x = 13 N/m * .04 m
-.52 N
F=kx=13N/m*.04m=.52 newtons
When an acrobat reaches the equilibrium position, the net force acting along the direction of motion is zero. Why does the acrobat swing past the equilibrium position?
Ah, the wonderful world of pinball machines! Let's spring into action and tackle this problem with humor and formulas.
To determine the force on the ball at the moment the spring is released, we can use Hooke's Law: F = kx, where F is the force, k is the spring constant, and x is the displacement (compression) of the spring.
In this case, the displacement (compression) of the spring is 4.0 cm, so x = 0.04 m (converting centimeters to meters).
Substituting the values into the formula, we have:
F = (13 N/m) x (0.04 m)
= 0.52 N
So, at the moment the spring is released, the force on the ball is 0.52 N.
As for your other questions, the ball leaves the spring when it's at the equilibrium position, meaning it has traveled past the compressed state and is ready to bounce off into pinball glory!
Regarding the ball's mass, it's not necessary for this particular calculation. Hooke's Law only requires the spring constant and displacement to determine the force.
I hope this explanation brightened up your understanding of the problem! If not, I can always juggle more information your way.
To solve this problem, you need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the spring
k is the spring constant
x is the displacement from the equilibrium position
In this case, the spring is compressed by 4.0 cm, so x = -0.04 m (negative because the displacement is in the opposite direction of the spring force). And the spring constant k is given as 13 N/m.
To find the force on the ball at the moment the spring is released, substitute the values into the formula:
F = -kx
F = -(13 N/m)(-0.04 m)
Performing the calculations:
F = 0.52 N
Therefore, the force on the ball at the moment the spring is released is 0.52 N.
As for your additional questions, the ball leaves the spring at the equilibrium position, which is 4 cm from compression. The equilibrium position is the position where the spring force and the inertial force on the ball are balanced. In this problem, you do not need to know the mass of the ball since you are only calculating the force exerted by the spring.