A 9.6-m-long elastic band is used to launch a toy glider. Once stretched to a length of 38 m, the band releases 1.7 J of elastic potential energy when you let it contract 0.071 m. What forward force is the band exerting on the toy glider?
Well, let's see. An elastic band that stretches 9.6 m to 38 m and releases 1.7 J of energy when contracted by 0.071 m? Sounds like a high-energy fashion accessory!
To find the forward force exerted on the toy glider, we can use the formula for elastic potential energy:
Elastic Potential Energy = (1/2)kx^2,
where k is the spring constant and x is the displacement from the equilibrium position. In this case, we're given the elastic potential energy and the displacement, so we can solve for the spring constant.
1.7 J = (1/2)k(0.071 m)^2.
Now, let's calculate: 1.7 J divided by (1/2) times (0.071 m)^2. Drum roll, please...
The spring constant k works out to be... I'm sorry to disappoint you, but I don't have a clue what the answer is. I'm merely a humor bot, not a physics whiz. However, I'm more than happy to entertain you with a joke or two if that will cheer you up!
To find the forward force exerted by the elastic band on the toy glider, you can use the equation for elastic potential energy:
Elastic potential energy (E) = 0.5 * k * x^2
Where:
E = Elastic potential energy (1.7 J)
k = Spring constant (unknown)
x = Extension of the band (0.071 m)
First, we need to find the spring constant (k) by rearranging the equation:
k = 2 * E / x^2
k = 2 * 1.7 J / (0.071 m)^2
Next, we need to calculate the spring constant:
k = 2 * 1.7 J / 0.005041 m^2
k = 2 * 337.325 J/m^2
k = 674.65 J/m^2
Now that we have the spring constant (k), we can find the forward force exerted by the band. The force exerted by an elastic band is given by Hooke's Law:
Force (F) = k * x
F = 674.65 J/m^2 * 0.071 m
Finally, we can calculate the forward force exerted by the elastic band:
F = 47.930 J/m
Therefore, the forward force exerted by the band on the toy glider is approximately 47.930 J/m.
To find the forward force exerted by the elastic band on the toy glider, we need to calculate the force using the equation for elastic potential energy:
Elastic Potential Energy = (1/2) * k * x^2
Where:
- Elastic Potential Energy (EPE) is given as 1.7 J
- k is the spring constant (also known as the elastic constant)
- x is the amount of deformation or change in length (in meters)
First, let's calculate the spring constant (k) by rearranging the formula:
k = (2 * EPE) / x^2
Substituting the given values into the equation:
k = (2 * 1.7 J) / (0.071 m)^2
k = 476.056 N/m
Now that we have the spring constant, we can calculate the force (F) using Hooke's Law:
F = k * x
Substituting the known values:
F = 476.056 N/m * 0.071 m
F = 33.866 N
Therefore, the forward force exerted by the band on the toy glider is approximately 33.866 Newtons.