Answer:
[tex]a=105\ m/s^2[/tex]
Explanation:
Spring Force
When a spring is compressed by a distance x, it exerts a force given by the law of Hooke:
[tex]F=k.x[/tex]
Where k is the constant of the spring. If we know this force and the mass m of the object launched by the spring, we can compute its acceleration by using Newton's formula
[tex]F=m.a[/tex]
[tex]\displaystyle a=\frac{F}{m}[/tex]
Replacing the formula for the spring
[tex]\displaystyle a=\frac{kx}{m}[/tex]
The values are
[tex]k=35 N/m\\m=100 gr=0.1 kg\\x=30 cm=0.3 m[/tex]
Thus
[tex]\displaystyle a=\frac{(35)(0.3)}{0.1}[/tex]
[tex]\boxed{a=105\ m/s^2}[/tex]
The toy rocket will have an acceleration of [tex]105\ m/s^2[/tex]
Note: If the launch would have been vertical, then the acceleration of gravity should have been used in the calculations
