A 2.1kg soccer ball is struck with a 45N force for .3 seconds. What is the Frictional force if the ball rolls for 11 seconds before stopping.
To calculate the frictional force, we first need to determine the initial velocity of the soccer ball. We can do this by using Newton's second law of motion.
According to Newton's second law, the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the acceleration is the change in velocity over time.
Since the soccer ball was struck with a force of 45N for 0.3 seconds, we can calculate the acceleration as follows:
Acceleration = Force / Mass
Acceleration = 45N / 2.1kg
Acceleration ≈ 21.43 m/s^2
Now, we need to determine the initial velocity of the ball. We can use the equation of motion:
Final Velocity = Initial Velocity + (Acceleration * Time)
Substituting the given values:
0 = Initial Velocity + (21.43 m/s^2 * 0.3s)
Simplifying the equation:
Initial Velocity = -21.43 m/s^2 * 0.3s
Initial Velocity = -6.43 m/s (negative sign indicates that the ball is slowing down)
Now, we can calculate the frictional force. The frictional force acting on a rolling object is given by the equation:
Frictional Force = Coefficient of Friction * Normal Force
However, we need to find the normal force acting on the soccer ball. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the ball.
Weight = Mass * Acceleration due to gravity
Weight = 2.1kg * 9.8 m/s^2
Weight ≈ 20.58N
Now that we have the weight, we can calculate the frictional force using the coefficient of friction. However, the coefficient of friction is not provided in the given information. To proceed, we need to assume a value for the coefficient of friction.
Let's assume a coefficient of friction of 0.3. Therefore:
Frictional Force = 0.3 * 20.58N
Frictional Force ≈ 6.18N
So, assuming a coefficient of friction of 0.3, the frictional force acting on the soccer ball is approximately 6.18N.