a projectile with mass (m)=35grams is placed on a vertically positioned spring. the spring constant (k)=6500N/m. the projectile is launched straight up into the air by compressing the spring a distance (x)=8cm.How high does the projectile go?
To determine how high the projectile goes, we can use the principles of conservation of mechanical energy. The energy stored in the spring is converted into the kinetic energy of the projectile when it is launched.
First, we need to calculate the potential energy stored in the spring when it is compressed a distance (x) = 8 cm. The potential energy stored in a spring is given by the formula:
Potential Energy (PE) = 0.5 * k * x^2
where k is the spring constant and x is the distance the spring is compressed.
Converting the given distance (x) from centimeters to meters:
x = 8 cm = 0.08 m
Substituting the values into the formula:
PE = 0.5 * 6500 N/m * (0.08 m)^2
= 0.5 * 6500 N/m * 0.0064 m^2
= 20.8 Joules
Next, we can equate the potential energy stored in the spring to the maximum potential energy of the projectile at its highest point, followed by solving for the height (h):
PE = m * g * h
where m is the mass of the projectile and g is the acceleration due to gravity.
Converting the given mass (m) from grams to kilograms:
m = 35 grams = 0.035 kg
Substituting the values into the formula:
20.8 J = 0.035 kg * 9.8 m/s^2 * h
Now, solving for h:
h = 20.8 J / (0.035 kg * 9.8 m/s^2)
≈ 58.67 m
Therefore, the projectile will reach a height of approximately 58.67 meters.