At an outdoor market, a bunch of bananas attached to the bottom of a vertical spring of force constant 16.0 N/m is set into oscillatory motion with an amplitude of 20.0 cm. It is observed that the maximum speed of the bunch of bananas is 39.6 cm/s. What is the weight of the bananas in newtons?
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Try using the fact that the maximum kinetic energy , (1/2)MVmax^2
is equal to the initial potential energy stored in the spring,
(1/2) k X^2.
Then solve for Vmax
k is the spring force constant, which you have been given. X is the starting (and maximum) deflection
To solve this problem, we can use the concept of simple harmonic motion and the equation for the maximum speed of an object undergoing simple harmonic motion:
v_max = Aω
Where v_max is the maximum speed, A is the amplitude, and ω is the angular frequency. The angular frequency can be expressed in terms of the force constant (k) and the mass (m) of the object:
ω = sqrt(k/m)
Given that the force constant is 16.0 N/m, and the amplitude is 20.0 cm, we can calculate the maximum speed by rearranging the equation:
v_max = Aω
v_max = 0.20 m * sqrt((16.0 N/m) / m)
To find the weight of the bananas, we need to convert the speed from cm/s to m/s. Since 1 m = 100 cm and 1 s = 1 s, we have:
v_max = (39.6 cm/s) * (1 m/100 cm) * (1 s/1 s)
v_max = 0.396 m/s
Now, we can solve the equation for the maximum speed to find the weight of the bananas. Rearranging the equation:
v_max = 0.20 m * sqrt((16.0 N/m) / m)
0.396 m/s = 0.20 m * sqrt((16.0 N/m) / m)
By squaring both sides:
(0.396 m/s)^2 = (0.20 m)^2 * ((16.0 N/m) / m)
Simplifying:
0.157104 m^2/s^2 = 0.04 m^2 * (16.0 N/m) / m
Canceling out the units of meter (m):
0.157104 m · s^(-2) = 0.04 m · (16.0 N/m)
Dividing both sides by 0.04 m:
(0.157104 m · s^(-2)) / (0.04 m) = 16.0 N
Simplifying:
3.9276 m · s^(-2) = 16.0 N
Therefore, the weight of the bananas in newtons is approximately 16.0 N.
To find the weight of the bananas in newtons, we need to use the equation:
Weight = mass × acceleration due to gravity
We can determine the mass of the bananas using the formula for the period of oscillation of a mass-spring system:
T = 2π√(m/k)
Where T is the period, m is the mass, and k is the force constant of the spring.
First, let's find the period. The period is the time taken for the bananas to complete one oscillation. In this case, we can take the maximum speed as the speed when the bananas are passing through the equilibrium position.
The maximum speed of the bananas is given as 39.6 cm/s. However, we need the speed in meters per second for consistent units. So, we convert it to m/s:
Maximum speed = 39.6 cm/s = 0.396 m/s
Next, let's find the period:
T = 2π / ω
where ω is the angular frequency, given by:
ω = 2πf
where f is the frequency of the oscillation. The frequency is the reciprocal of the period:
f = 1 / T
Let's calculate the frequency and then find the period:
f = 1 / T = 1 / (2π / ω) = ω / (2π) = (0.396 m/s) / (2π) ≈ 0.063 Hz
T = 1 / f ≈ 1 / 0.063 Hz ≈ 15.873 sec
Now, let's find the mass of the bananas. To do that, we use the equation:
T = 2π√(m/k)
Squaring both sides to get rid of the square root:
T^2 = 4π^2(m/k)
Rearranging the equation to isolate m:
m = (T^2 × k) / (4π^2)
Plugging in the values:
m = (15.873 sec)^2 × (16.0 N/m) / (4π^2)
m ≈ 1.593 kg
Finally, we can calculate the weight:
Weight = mass × acceleration due to gravity
Weight ≈ (1.593 kg) × (9.8 m/s^2)
Weight ≈ 15.6 N
Therefore, the weight of the bunch of bananas is approximately 15.6 newtons.