What is the speed of a hockey puck of mass 0.15 kg if it experiences a force 60 in N for the extent of a slap shot of 0.25 m?
F=ma
a=F/m
s=v²/2a
v=sqrt(2as) = sqrt(2Fs/m)
.05
To find the speed of the hockey puck, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration (F = m * a).
In this case, we are given the force (F = 60 N) and the mass of the puck (m = 0.15 kg). We need to find the acceleration of the puck.
First, we rearrange the equation to solve for acceleration:
a = F / m
Now, we can substitute the values:
a = 60 N / 0.15 kg
a ≈ 400 m/s²
Next, we can use the kinematic equation to find the final velocity (v) of the puck, given the initial velocity (u = 0 m/s), acceleration (a), and distance (s = 0.25 m):
v² = u² + 2as
Since the initial velocity is 0 m/s:
v² = 0² + 2 * 400 m/s² * 0.25 m
v² = 2 * 400 m²/s² * 0.25 m
v² = 200 m²/s² * 0.25 m
v² = 50 m²/s² * m
Taking the square root of both sides to solve for v:
v ≈ sqrt(50) m/s
v ≈ 7.07 m/s
Therefore, the speed of the hockey puck after experiencing a force of 60 N for a slap shot of 0.25 m is approximately 7.07 m/s.