A hockey puck has an initial velocity of 44 m/s and a final velocity of 34 m/s.
(a) If it travels 25 m during this time, what is the acceleration?
m/s2
(b) If the mass of the puck is 0.13 kg, what is the horizontal force on the puck?
N
a) Vf^2=Vi^2+2ad solve for a.
b) force=mass*accelearation
To find the acceleration, we can use the following equation:
vf^2 = vi^2 + 2ad
Where:
vf = final velocity = 34 m/s
vi = initial velocity = 44 m/s
d = displacement = 25 m
a = acceleration (to be found)
(a) Rearranging the equation, we get:
a = (vf^2 - vi^2) / (2d)
Substituting the given values:
a = (34^2 - 44^2) / (2 * 25) = (-400) / 50 = -8 m/s^2
Therefore, the acceleration of the hockey puck is -8 m/s^2.
(b) To find the force, we can use Newton's second law of motion:
F = ma
Where:
F = force (to be found)
m = mass of the puck = 0.13 kg
a = acceleration = -8 m/s^2 (as calculated in part a)
Substituting the given values:
F = 0.13 kg * (-8 m/s^2) = -1.04 N
Therefore, the horizontal force on the puck is -1.04 N.
To find the acceleration of the hockey puck, we can use the formula:
Acceleration = (Final velocity - Initial velocity) / Time
However, we don't have the time in this case, but we do have the distance traveled. We can use the formula:
Distance = (Initial velocity + Final velocity) / 2 * Time
Rearranging this formula, we can solve for time:
Time = 2 * Distance / (Initial velocity + Final velocity)
Substituting the given values, we can find the time:
Time = 2 * 25 m / (44 m/s + 34 m/s)
Time = 2 * 25 m / 78 m/s
Time ≈ 0.641 s
Now that we have the time, we can use it to find the acceleration:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (34 m/s - 44 m/s) / 0.641 s
Acceleration = -10 m/s / 0.641 s
Acceleration ≈ -15.591 m/s²
Since acceleration is a vector quantity, the negative sign indicates that the puck is decelerating or slowing down.
Now let's move on to the second question to find the horizontal force on the puck. We can use Newton's second law of motion:
Force = Mass * Acceleration
Substituting the values:
Force = 0.13 kg * (-15.591 m/s²)
Force ≈ -2.03 N
Again, the negative sign indicates that the force is acting in the opposite direction of the displacement. In this case, it means that the force is acting to slow down the puck.