A 68.5kg skater moving initially at 2.40m/s on rough horizontal ice comes to rest uniformly in 3.52s due to friction from the ice. I need help finding the force friction exert on the skater.
Use the rule that impulse = momentum change.
Friction force * (time applied) = loss of linear momentium
F * 3.52 s = (68.5 kg)*(2.40 m/s)
Solve for F, which will be in Newtons
Thank you
Use average acceleration equation to find the acceleration: a=delta(v)/delta(t), then plug into the newtons 2nd law of motion, which is the equation vector form.
To find the force of friction exerted on the skater, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
In this case, the skater comes to rest uniformly in 3.52 seconds, which means the acceleration is constant and equal to zero. Since the initial velocity is 2.40 m/s and the final velocity is 0 m/s, we can use the equation of motion:
Final velocity (v) = Initial velocity (u) + Acceleration (a) × Time (t)
Rearranging the equation, we can find the acceleration:
Acceleration (a) = (Final velocity - Initial velocity) / Time
a = (0 - 2.40) m/s / 3.52 s
a = -2.40 m/s / 3.52 s
Now, we know the acceleration of the skater, so we can calculate the force of friction using Newton's second law:
Force of friction (F) = Mass (m) × Acceleration (a)
F = 68.5 kg × (-2.40 m/s / 3.52 s)
Calculating the force of friction, we get:
F ≈ -46.50 N
The negative sign indicates that the force of friction is acting in the opposite direction to the skater's motion, causing the skater to slow down.