A ball of mass m 5
1.80 kg is released from
rest at a height h 5
65.0 cm above a light
vertical spring of force
constant k as in Figure
P5.66a. The ball strikes
the top of the spring
and compresses it a
distance d 5 9.00 cm
as in Figure P5.66b.
Neglecting any energy
losses during the collision, find (a) the speed of the
ball just as it touches the spring and (b) the force con-
stant of the spring.
M = 1.80 kg.
h = 65 cm. = 0.65 m.
a. V^2 = Vo^2 + 2g*h. = 0 + 19.6*0.65 = 12.74.
V = 3.57 m/s.
k=(mgh)/(1/2)(x^2)
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m
A) v= square root of: ((2)(g)(h)) => square root of (2*9.8*0.65)
= 3.57 m/s
B) k= (mgh)/(1/2)(x^2)
k=(1.8)(9.8)(0.09+0.65)/(1/2)(0.09^2)
k=3223.1 N/m / 1000
= 3.223 kN/m
(a) Well, before we calculate the speed of the ball, let's take a moment to appreciate the bravery of that little ball. It must have a lot of courage to go barreling towards that spring like that. Anyway, back to business.
To find the speed of the ball just as it touches the spring, we can use the conservation of energy. Since there are no energy losses, the potential energy of the ball at its initial position will be equal to the sum of its kinetic energy and potential energy at the final position.
The potential energy at the initial position is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height. Plugging in the values, we get potential energy = (1.80 kg)(9.8 m/s^2)(0.65 m).
At the final position, the potential energy is zero since the ball is compressed against the spring. Therefore, the kinetic energy is equal to the initial potential energy.
Setting these two equal, we get (1/2)m(v^2) = mgh. Canceling out the mass and solving for v, we find the speed of the ball just as it touches the spring.
Now, if only we could bottle up that kind of energy and enthusiasm, the world would be a much happier place.
(b) To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement. In this case, the displacement is given as 9.00 cm.
So, the force exerted by the spring is F = kx, where F is the force, k is the force constant, and x is the displacement.
We can also calculate the force by using Newton's second law, which states that the force is equal to the mass times the acceleration. In this case, the acceleration is the change in velocity of the ball divided by the time it takes to compress the spring. Since the collision happens very quickly, we can assume that the time is negligible.
So, we have F = ma, where F is the force, m is the mass of the ball, and a is the acceleration.
Setting these two equations equal, we get kx = ma. Canceling out the mass, we find the force constant of the spring.
So, there you go! The speed of the ball just as it touches the spring and the force constant of the spring. Now, go out there and bounce back from any challenge just like that little ball.
To find the speed of the ball just as it touches the spring and the force constant of the spring, we can apply the principle of conservation of mechanical energy.
(a) The speed of the ball just as it touches the spring can be found using the conservation of mechanical energy equation:
mgh = (1/2)mv^2
where m is the mass of the ball, g is the acceleration due to gravity, h is the initial height of the ball, and v is the velocity of the ball just as it touches the spring.
In this case, m = 1.80 kg, g = 9.8 m/s^2, and h = 65.0 cm = 0.65 m. Solving for v, we have:
0.5 * 1.80 kg * 9.8 m/s^2 * 0.65 m = (1/2) * 1.80 kg * v^2
v^2 = (0.5 * 1.80 kg * 9.8 m/s^2 * 0.65 m) / (1/2 * 1.80 kg)
v^2 = (8.993 J) / 0.9
v^2 = 9.992 J
Taking the square root of both sides, we find:
v = sqrt(9.992 J)
v ≈ 3.16 m/s
Therefore, the speed of the ball just as it touches the spring is approximately 3.16 m/s.
(b) The force constant of the spring, represented by k, can be determined using Hooke's Law:
F = -k * x
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
In this case, the displacement of the spring is 9.00 cm = 0.09 m, and the force exerted by the spring can be calculated using the formula for potential energy:
PE = (1/2)kx^2
The potential energy stored in the compressed spring must be equal to the potential energy lost by the ball as it falls:
mgh = (1/2)kx^2
Rearranging the equation for k, we have:
k = (2mgh) / x^2
Substituting the given values, we get:
k = (2 * 1.80 kg * 9.8 m/s^2 * 0.65 m) / (0.09 m)^2
k = (22.176 J) / 0.0081 m^2
k ≈ 2737.78 N/m
Therefore, the force constant of the spring is approximately 2737.78 N/m.