A slingshot with k = 400. N/m is pulled back 0.0500 m to shoot a 0.100-kg stone. If the slingshot projects the stone straight up in the air, (a) what is the speed when the stone is first released, and (b) what is the maximum height to which the stone will rise? [neglect friction]
potential energy = (1/2)(400)(.5)^2 = 50 J
so Ke = 50 = (1/2) m v^2
(1/2)(.1)v^2 = 50
v = 31.6 m/s
potential energy at top = m g h = 50
.1 * 9.8 * h = 50
h = 51 meters
(a) (Stored slingshot potential energy) = (Kinetic energy of stone when released)
(1/2)kX^2 = (1/2)MVo^2
Vo = X*sqrt(k/M)
= 0.05*sqrt(400/0.1) = 3.16 m/s
(which is not very fast)
(b) (Maximum gravitational potential energy) = (Stored slingshot potential energy)
M g H = (1/2) k X^2
H = [k/(2Mg)]*X^2
(not very high, either)
Are you sure the pull back is only X = 0.05 meters?
Thanks so much! Yeah, it's weird it says 0.05 meters. Thanks though!
To solve this problem, we can use the principles of work and energy. The potential energy stored in the slingshot is converted into kinetic energy of the stone when it is released.
(a) To find the speed when the stone is first released, we can use the concept of potential energy. The potential energy stored in the slingshot is given by the formula:
Potential energy = (1/2) * k * x^2
Where:
k = spring constant = 400 N/m
x = distance the slingshot is pulled back = 0.0500 m
Substituting these values into the formula:
Potential energy = (1/2) * 400 N/m * (0.0500 m)^2
= 0.5 * 400 N/m * 0.0025 m^2
= 0.5 * 1 N * 0.0025 m^2
= 0.00125 J
Now, we know that this potential energy is converted into the kinetic energy of the stone when it is released:
Potential energy = Kinetic energy
0.00125 J = (1/2) * mass * velocity^2
Rearranging the equation to solve for velocity:
velocity^2 = (2 * Potential energy) / mass
velocity^2 = (2 * 0.00125 J) / 0.100 kg
velocity^2 = 0.025 J / 0.100 kg
velocity^2 = 0.25 m^2/s^2
Taking the square root of both sides to find the velocity:
velocity = sqrt(0.25 m^2/s^2)
velocity = 0.5 m/s
Therefore, the speed when the stone is first released is 0.5 m/s.
(b) To find the maximum height to which the stone will rise, we need to consider the conservation of mechanical energy. At the maximum height, all the initial kinetic energy of the stone will be converted into potential energy.
The total mechanical energy of the system remains constant throughout the motion. Therefore, at the maximum height:
Kinetic energy + Potential energy = Constant
Initially, the stone was released with a kinetic energy of (1/2) * mass * velocity^2. At the maximum height, the stone comes to rest, so its kinetic energy becomes zero. Therefore:
Potential energy at maximum height = (1/2) * mass * velocity^2
Substituting the given values:
Potential energy at maximum height = (1/2) * 0.100 kg * (0.5 m/s)^2
= (1/2) * 0.100 kg * 0.25 m^2/s^2
= 0.0125 J
Since the potential energy at maximum height is equal to the gravitational potential energy, we can use the formula:
Potential energy = mass * g * height
Where:
g = acceleration due to gravity = 9.8 m/s^2 (approximate value)
Rearranging the formula to solve for height:
height = Potential energy / (mass * g)
height = 0.0125 J / (0.100 kg * 9.8 m/s^2)
height = 0.0125 J / 0.980 kg*m^2/s^2
height = 0.0128 m (rounded to four significant figures)
Therefore, the maximum height to which the stone will rise is approximately 0.0128 meters.