An ice skater of mass 50kg moving at 10 m/s coasts to a halt in 100 m on a smooth ice surface
a. calculater acceleration of the skater (.5 m/s)
b. what is the force of friction between the ice and the skates? (25 N)
c. calculate the coefficient of friction (.051)
are my answers correct?
yes.
thanks!
no
To verify if your answers are correct, let's go through each question one by one:
a. To calculate the acceleration of the skater, we can use the kinematic equation:
`v^2 = u^2 + 2as`
where:
v = final velocity (0 m/s, since the skater comes to a halt)
u = initial velocity (10 m/s)
a = acceleration (unknown)
s = distance (100 m)
Rearranging the equation, we have:
`a = (v^2 - u^2) / (2s)`
`a = (0^2 - 10^2) / (2 * 100)`
`a = -100 / 200`
Therefore, the acceleration of the skater is -0.5 m/s² (since the skater is slowing down, the acceleration is negative).
So, your answer of -0.5 m/s² is correct.
b. To calculate the force of friction between the ice and the skates, we can use Newton's second law:
`F = ma`
where:
F = force of friction (unknown)
m = mass of the skater (50 kg)
a = acceleration (calculated as -0.5 m/s²)
Plugging in the values, we have:
`F = 50 kg * (-0.5 m/s²)`
`F = -25 N` (Note: the negative sign indicates that the force is acting in the opposite direction of motion)
Therefore, the force of friction between the ice and the skates is -25 N.
So, your answer of -25 N is correct.
c. To calculate the coefficient of friction, we can use the equation:
`F = μ * N`
where:
F = force of friction (25 N)
μ = coefficient of friction (unknown)
N = normal force (equal to the weight of the skater, which is 50 kg * 9.8 m/s²)
Rearranging the equation, we have:
`μ = F / N`
`μ = 25 N / (50 kg * 9.8 m/s²)`
`μ ≈ 0.051`
Therefore, the coefficient of friction between the ice and the skates is approximately 0.051.
Your answer of approximately 0.051 is correct.
In conclusion, all of your answers (a, b, and c) are correct. Well done!