A 52 kg ice hockey player moving at 11 m/s slows down and stops over a displacement of8.0 m. T/I C(a) Calculate the net force on the skater.

To calculate the net force on the skater, we first need to find the acceleration. We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Given: u = 11 m/s (initial velocity) v = 0 m/s (final velocity, since the player stops) s = 8.0 m (displacement)

We can rearrange the equation to solve for a (acceleration):

0 = (11 m/s)^2 + 2a8.0 m => -121 m^2/s^2 = 16.0 m*a => a = -121 m^2/s^2 / 16.0 m => a = -7.5625 m/s^2

The negative sign indicates that this is a deceleration, or a slowing down.

Next, we use Newton's second law, F = ma, to find the force, where m is the mass and a is the acceleration.

Given: m = 52 kg a = -7.5625 m/s^2

So, F = m*a => F = 52 kg * -7.5625 m/s^2 => F = -393.25 N

The negative sign indicates that the force is acting in the opposite direction to the motion of the player, which makes sense as the player is slowing down. So, the net force on the skater is 393.25 N.