A pinball machine launches a 139 g ball
with a spring-driven plunger. The game
board is inclined at 8 ◦
above the horizontal.
Assume: The plunger’s mass and frictional
effects are negligible. The acceleration of
gravity is 9.8 m/s2
Find the force constant k of the spring that
will give the ball a speed of 123 cm/s when the
plunger is released from rest with the spring
compressed 5 cm from its relaxed position.
Answer in units of N/m.
To find the force constant of the spring (k), we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its 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 force constant (which we want to find), and x is the displacement from the equilibrium position.
In this case, the spring is compressed by 5 cm from its relaxed position, so x = -0.05 m (negative sign indicates compression).
We want to find the force constant (k) that will give the ball a speed of 123 cm/s.
The work done by the spring is given by:
W = (1/2)kx^2
The work done by the spring is equal to the kinetic energy of the ball:
W = (1/2)mv^2
Where m is the mass of the ball and v is its velocity. The mass of the ball is given as 139 g, but we need to convert it to kilograms, so m = 0.139 kg. The velocity of the ball is given as 123 cm/s, so v = 1.23 m/s.
Setting the equations for work equal to each other:
(1/2)kx^2 = (1/2)mv^2
Simplifying:
kx^2 = mv^2
k = (mv^2) / x^2
Substituting the given values:
k = (0.139 kg * (1.23 m/s)^2) / (-0.05 m)^2
Calculating:
k ≈ 71.1 N/m
Therefore, the force constant k of the spring that will give the ball a speed of 123 cm/s when the plunger is released is approximately 71.1 N/m.