A 22.6-N horizontal force is applied to a 0.0710-kg hockey puck to accelerate it across the ice from an initial rest position. Ignore friction and determine the final speed (in m/s rounded off to a whole number) of the puck after being pushed for a time of 0.0721 second
Upon neglecting air resistance, there are three forces acting upon the object. The up and down force balance each other and the acceleration is caused by the applied force. The net force is 22.6 N, right (equal to the only rightward force - the applied force). So the acceleration of the object can be computed using Newton's second law.
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a = Fnet / m = (22.6 N, right) / (0.0710 kg) = 318 m/s/s, right
The acceleration value can be used with other kinematic information (vi = 0 m/s, t = 0.0721 s) to calculate the final speed of the puck. The kinematic equation, substitution and algebra steps are shown.
vf = vi + a•t
vf = 0 m/s + (318 m/s/s)•(0.0721 s)
vf = 23.0 m/s
Well, well, well, looks like we have a high-speed hockey puck on our hands! Let's see how fast this slippery little fella can go.
We can use Newton's second law of motion, which states that the force on an object is equal to its mass times its acceleration (F = ma). Rearranging the equation, we get a = F/m.
In this case, the force applied to the puck is 22.6 N, and its mass is 0.0710 kg. Plugging those values in, we get:
a = 22.6 N / 0.0710 kg ≈ 318.3 m/s²
Now, we know acceleration, and we also know the time for which the force is applied, which is 0.0721 seconds. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since the puck starts from rest), a is the acceleration, and t is time.
Plugging in the numbers, we get:
v = 0 + (318.3 m/s²)(0.0721 s) ≈ 22.97 m/s
So, after being pushed for a brief moment in time, our speedy little puck will have a final speed of approximately 23 m/s. That's enough to make any hockey player jealous!
To find the final speed of the puck, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass. The formula is:
F = m * a
Where:
F = Net force applied (22.6 N)
m = Mass of the object (0.0710 kg)
a = Acceleration of the object
Rearranging the formula, we get:
a = F / m
Now, we need to find the acceleration of the puck.
Substituting the given values into the formula:
a = 22.6 N / 0.0710 kg
a ≈ 318.3 m/s^2
Next, we can use the equation for constant acceleration to find the final speed of the object. The equation is:
v = u + a * t
Where:
v = Final velocity (unknown)
u = Initial velocity (0 m/s as the puck starts from rest)
a = Acceleration (318.3 m/s^2)
t = Time taken (0.0721 s)
Substituting the values:
v = 0 m/s + 318.3 m/s^2 * 0.0721 s
v ≈ 22.98 m/s
Rounding off to a whole number, the final speed of the hockey puck after being pushed for a time of 0.0721 seconds is approximately 23 m/s.
To determine the final speed of the hockey puck, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be represented as:
F = ma
Where:
F = Net force applied (given as 22.6 N)
m = Mass of the hockey puck (given as 0.0710 kg)
a = Acceleration of the hockey puck
We can rearrange the equation to solve for acceleration:
a = F/m
Substituting the given values:
a = 22.6 N / 0.0710 kg
Now, we need to determine the final velocity of the puck after being pushed for a specific time. To do this, we can use the kinematic equation:
v = u + at
Where:
v = Final velocity (unknown)
u = Initial velocity (since the puck starts from rest, u = 0)
a = Acceleration (already calculated)
t = Time (given as 0.0721 s)
Rearranging the equation gives us:
v = at
Substituting the known values:
v = (22.6 N / 0.0710 kg) * 0.0721 s
Calculating this will give us the final velocity of the puck in m/s.