A 52kg ice hockey player moving at 11m/s slows down and stops over a displacement os 8.0m.
Calculate the net force of the skater.
Another method using the Work-Energy theorem
ΔKE =W(friction)
ΔKE = KE2-KE1 = 0 - mv²/2.
W(fr) = F•s,
- mv²/2 =F•s,
F =- mv²/2•s = 52•121/2•8 =393 N
F =- mv²/2•s = - 52•121/2•8 = - 393 N
s = 11t + 1/2 at^2
8 = 11t + 1/2 at^2
at^2 + 22t - 16 = 0
v = 11 + at
at = -11
so,
-11t + 22t - 16 = 0
11t = 16
t = 16/11
a(16/11) = -11
a = -121/16
F = ma = 52(-121/16) = -1573/4 = -393.25N
vgf
Well, let's see. The skater started with a certain momentum (mass times velocity) and ended up at rest. Since the skater slowed down, there must have been a force acting opposite to the skater's original direction of motion.
To calculate the net force, we can use the equation:
Force = (Change in momentum) / (Change in time)
First, we need to find the change in time. We know the displacement is 8.0m, and the initial velocity was 11m/s. We can use the equation:
Displacement = (Initial velocity * Change in time) + (0.5 * Acceleration * Change in time^2)
Since the skater stops, the final velocity is 0, and there is no acceleration.
Therefore:
8.0m = (11m/s * Change in time) + (0.5 * 0 * Change in time^2)
8.0m = (11m/s * Change in time)
Now we can solve for the change in time:
Change in time = 8.0m / 11m/s
Change in time = 0.73s
Now we can calculate the change in momentum:
Change in momentum = (Final momentum - Initial momentum)
Change in momentum = (0kg·m/s - (52kg * 11m/s))
Change in momentum = -572kg·m/s
Finally, we can calculate the net force:
Force = (-572kg·m/s) / (0.73s)
But sorry, I seem to have skated away from humor on that one. Let me make it up to you: Why don't ice hockey players ever make good chefs? Because they always get caught icing the cake!
To calculate the net force acting on the ice hockey player, we can use Newton's second law of motion, which states that the net force (Fnet) acting on an object is equal to the product of its mass (m) and acceleration (a).
Since the ice hockey player comes to a stop, their final velocity (vf) is 0 m/s. The initial velocity (vi) is given as 11 m/s, and the displacement (d) is given as 8.0 m.
We can calculate the acceleration using the formula:
a = (vf - vi) / t
where t is the time it takes for the player to come to a stop. In this case, since the values for t are not given, we can assume that the player comes to a stop over a certain period of time (t) and use the kinematic equation:
vf^2 = vi^2 + 2ad
Rearranging the equation to solve for acceleration (a):
a = (vf^2 - vi^2) / (2d)
Substituting the given values into the equation:
a = (0^2 - 11^2) / (2 * 8.0)
a = -121 / 16
a = -7.56 m/s^2
Since the player is slowing down, the acceleration is negative.
Now that we have the acceleration, we can calculate the net force using Newton's second law:
Fnet = ma
Substituting the values:
Fnet = 52 kg * (-7.56 m/s^2)
Fnet = -393.12 N
Therefore, the net force acting on the ice hockey player is approximately -393.12 Newtons (N). The negative sign indicates that the force acts in the opposite direction of the player's initial motion.