A hockey player shoots the puck which is initially at rest by applying a force of 72 N through a displacement of 1.6 m. The mass of the puck is 161 grams.
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A hockey player take a slapshot on a puck at rest weighing 0.165 kg. Puck contact is for 0.038 s, Puck velocity is 48.64 m/s.
a) What is force acting on puck?
b) How far will puck travel if co-efficient of static friction with ice is 0.06?
To find the final velocity of the puck shot by the hockey player, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the force acting on the puck is 72 N.
First, let's convert the mass of the puck from grams to kilograms. There are 1000 grams in a kilogram, so the mass of the puck is 161 grams / 1000 = 0.161 kilograms.
Next, we can use the formula:
Force = mass * acceleration
Rearranging the formula, we get:
Acceleration = Force / mass
Plugging in the values, we have:
Acceleration = 72 N / 0.161 kg
Now we have the acceleration of the puck. We can also calculate the displacement using the formula:
Displacement = initial velocity * time + (0.5 * acceleration * time^2)
Since the puck is initially at rest, the initial velocity is 0 m/s. Therefore, the formula simplifies to:
Displacement = (0.5 * acceleration * time^2)
We know the displacement is 1.6 m, so we can rearrange the formula to solve for time:
Time^2 = (2 * Displacement) / acceleration
Plugging in the values, we have:
Time^2 = (2 * 1.6 m) / (72 N / 0.161 kg)
Finally, we can find the final velocity by using the formula:
Final velocity = initial velocity + (acceleration * time)
Since the initial velocity is 0 m/s, the formula simplifies to:
Final velocity = acceleration * time
Plugging in the values, we have:
Final velocity = (72 N / 0.161 kg) * sqrt((2 * 1.6 m) / (72 N / 0.161 kg))
Now you can calculate the final velocity of the puck by plugging in the values and solving the equation.