A flat stone is thrown horizontally across a frozen lake. If the stone decelerates at 2.5 meter per second squared, what is the coefficient of friction between the stone and the ice?take g equals 10 meter per second
I did this already
mu m g = m a = 2.5 m
mu = 2.5/g = 2.5/10 = 0.25
To find the coefficient of friction between the stone and the ice, we need to use the equation that relates the deceleration of the stone to the coefficient of friction and the acceleration due to gravity.
The equation for deceleration is given by:
deceleration = acceleration due to gravity - frictional force
Rearranging the equation, we get:
frictional force = acceleration due to gravity - deceleration
Given that the acceleration due to gravity (g) is 10 meters per second squared, and the deceleration is 2.5 meters per second squared, we can plug these values into the equation:
frictional force = 10 m/s^2 - 2.5 m/s^2
Simplifying, we get:
frictional force = 7.5 m/s^2
Now, the frictional force can be expressed as the product of the coefficient of friction (μ) and the normal force (N). Since the stone is thrown horizontally, the normal force is equal to the weight of the stone.
The weight of the stone (N) can be calculated using the equation:
weight (N) = mass (m) * acceleration due to gravity (g)
Since the mass of the stone is not given in the question, we cannot determine the coefficient of friction accurately without this information.
To solve for the coefficient of friction (μ), we would need to know the mass of the stone. Once we have the mass, we can calculate the weight (N) and equate it to the frictional force (7.5 m/s^2), allowing us to find the coefficient of friction (μ).