A player strikes a hockey puck giving it a velocity of 47.673 m/s. The puck slides across the ice for 0.175 s after which time its velocity is 46.473 m/s.
The acceleration of gravity is 9.8 m/s2 . If the puck strikes the goalie’s pads and stops in a distance of 4.39 cm, what average force is exerted on the pads?
Answer in units of N.
Sorry, I am not interested in the original speed or gravity or how long it slid across the ice.
However I do need the mass of the puck.
F = m *(change in velocity)/time
time = distance/average speed stopping
distance = .0439m
average speed stopping = 23.236 m/s
time = .0439/23.236 seconds to stop
F = m (46.473) (23.236/.0439)
so if I only knew m, I could help you.
However as you can sense, the heavier the puck, the more the force.
The mass of the hockey puck is 185 grams. When you said time = distance/average speed stopping, in the equation F = m (46.473)*(23.236/.0439), did you mean this, .0439/23.236.
To find the average force exerted on the pads, we can use the equation:
\[F = \frac{mv}{t}\]
where F is the force, m is the mass of the puck, v is the change in velocity, and t is the time.
First, let's find the change in velocity:
\[v = v_f - v_i = 46.473 m/s - 47.673 m/s = -1.2 m/s\]
The negative sign indicates that the velocity is decreasing.
Next, let's convert the distance traveled by the puck to meters:
\[d = 4.39 cm = 0.0439 m\]
Now, we can use the equation:
\[a = \frac{2d}{t^2}\]
where a is the acceleration. Since the puck is stopping, the final velocity is 0 m/s, so the equation can be rearranged to solve for the time:
\[t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 0.0439 m}{-1.2 m/s}} \approx 0.065125 s\]
Now, let's substitute the values into the formula for force:
\[F = \frac{m \times v}{t} = \frac{m \times (-1.2 m/s)}{0.065125 s} = -18.48 m \times \frac{kg}{s}\]
Finally, we can convert the force from kg/s to Newtons:
\[1 \frac{kg}{s} = 1 N\]
So, the average force exerted on the pads is approximately -18.48 N. The negative sign indicates that the force is in the opposite direction of the initial velocity of the puck.
To find the average force exerted on the goalie's pads, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a), or F = m * a.
First, let's calculate the acceleration of the puck. We can use the formula for acceleration, a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time interval.
Given:
- Initial velocity (v_i) = 47.673 m/s
- Final velocity (v_f) = 46.473 m/s
- Time interval (t) = 0.175 s
Substituting the values into the formula, we get:
a = (46.473 - 47.673) / 0.175
a = -1.2 / 0.175
a ≈ -6.857 m/s^2
The acceleration is negative because it represents the deceleration of the puck.
Next, let's convert the distance the puck stops to meters. The distance is given as 4.39 cm, so we divide it by 100 to convert it to meters:
Distance (d) = 4.39 cm / 100
d = 0.0439 m
Now, we need to calculate the mass of the puck. We can use the formula for weight, W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
Given:
- Acceleration due to gravity (g) = 9.8 m/s^2
Rearranging the formula, we have:
m = W / g
Since the weight is not given, we need to use the distance stopped (d) to find it. The work done (W) is equal to force (F) times distance (d), or W = F * d. Rearranging this formula, we have:
F = W / d
Since the puck comes to a stop, all of its kinetic energy is transformed into work (W) done by the force exerted on the pads. The initial kinetic energy (KE_i) is given by:
KE_i = 0.5 * m * v_i^2
Since the puck stops, its final kinetic energy (KE_f) is zero.
Hence, the work (W) done by the force can be calculated as the difference between the initial and final kinetic energies:
W = KE_i - KE_f
W = 0.5 * m * v_i^2 - 0
Substituting in the given values, we have:
W = 0.5 * m * (47.673 m/s)^2
Now, we can substitute the value of W into the formula for F to find the average force exerted on the pads:
F = W / d
F ≈ (0.5 * m * (47.673 m/s)^2) / 0.0439 m
F ≈ (0.5 * m * 2276.282329 m^2/s^2) / 0.0439 m
F ≈ (1138.141 m * kg/s^2) / 0.0439 m
F ≈ 25932.285 N
Therefore, the average force exerted on the goalie's pads is approximately 25932.285 N.