a hockey puck is given the initial speed of 10 m/s. If the coefficient of kinetic friction between the ice and puck is 0.10, how far will the puck slide before stopping?
What you wanna do is snort a line, smoke a J and look beyond what you see
f = ma
u(k) * f(normal) = m*a
u(k) *m*g=m*a
u(k) *g =a
a = 0.10 * 9.81m/s^2 = 0.981 m/s^2
this is defined to be a negative acceleration
v = u + a*t
so t = (v-u)/a
and x = t * (v+u)/2
so together x = (v+u)(v-u)/2a
where x = distance, v = final speed, u = initial speed, t = time, a = acceleration
substitute in our known values:
x = (0+10)(0-10)/(2*-0.981)
x = 50/0.981 = 50.97m
To determine how far the puck will slide before stopping, we need to use the equations of motion.
First, let's calculate the acceleration due to the kinetic friction. The force of kinetic friction can be found using the equation:
\(F_{\text{friction}} = \mu \times F_{\text{normal}}\)
where \(F_{\text{normal}}\) is the normal force and \(\mu\) is the coefficient of kinetic friction. The normal force is equal to the weight of the puck, which can be found using the equation:
\(F_{\text{normal}} = m \times g\)
where \(m\) is the mass of the puck and \(g\) is the acceleration due to gravity (approximately 9.8 m/s²).
Now, we can substitute the known values into the equations. Let's assume the mass of the puck is 0.1 kg (this value can be changed if necessary):
\(F_{\text{normal}} = m \times g = 0.1 \, \text{kg} \times 9.8 \, \text{m/s²} = 0.98 \, \text{N}\)
Next, we calculate the force of friction:
\(F_{\text{friction}} = \mu \times F_{\text{normal}} = 0.10 \times 0.98 \, \text{N} = 0.098 \, \text{N}\)
Now we know the force acting against the motion of the puck. We can use Newton's second law of motion, \(F = m \times a\), to find the acceleration of the puck. In this case, the force is the force of friction, and the mass is the mass of the puck:
\(F = m \times a\)
\(0.098 \, \text{N} = 0.1 \, \text{kg} \times a\)
Solving for \(a\), we find:
\(a = \frac{0.098 \, \text{N}}{0.1 \, \text{kg}} = 0.98 \, \text{m/s²}\)
Now, we can use the equation of motion for uniformly accelerated motion:
\(v^2 = u^2 + 2a \cdot s\)
where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement.
In this case, the final velocity (\(v\)) is 0 m/s because the puck stops, the initial velocity (\(u\)) is 10 m/s, the acceleration (\(a\)) is -0.98 m/s² (negative value because it's acting against the direction of motion), and we want to find the displacement (\(s\)).
Substituting the known values into the equation:
\(0^2 = 10^2 + 2 \cdot (-0.98) \cdot s\)
Simplifying the equation:
\(0 = 100 - 1.96s\)
Rearranging the equation to solve for \(s\):
\(1.96s = 100\)
\(s = \frac{100}{1.96} \approx 51.02 \, \text{m}\)
Therefore, the puck will slide approximately 51.02 meters before stopping.
KE= energy in friction
1/2 m v^2=mu*mg*distance
solve for distance.