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?
To determine the forward force exerted by the elastic band on the toy glider, we need to use the concept of elastic potential energy and Hooke's Law.
First, let's calculate the initial potential energy of the elastic band when it is stretched to a length of 38 m.
Elastic Potential Energy (PE) is given by the formula:
PE = 0.5 * k * x^2
Where:
PE is the elastic potential energy
k is the spring constant (stiffness) of the elastic band
x is the extension from the equilibrium position of the band
We are given that the band releases 1.7 J of elastic potential energy when it contracts by 0.071 m. So, we have the equation:
1.7 J = 0.5 * k * (0.071 m)^2
Now, let's solve this equation to find the spring constant (k):
k = (1.7 J) / [0.5 * (0.071 m)^2]
Next, we can determine the stretch (extension) of the band by subtracting the original length from the contracted length:
Stretch = 38 m - 9.6 m = 28.4 m
Now, we can find the force exerted by the elastic band using Hooke's Law:
F = k * x
Where:
F is the force exerted by the elastic band
k is the spring constant we calculated earlier
x is the stretch (extension) of the band
Finally, substituting the values into the equation, we get:
F = (k) * (28.4 m)
Now, plug in the value of k that we calculated earlier to find the force exerted by the elastic band on the toy glider.