Question

1. curling is a sport where competitors slide special stones across the ice, aiming for certain targets. janessa is in a curling competition. her stone has a mass of 18.9 kg, and the coefficient of friction between the stone and the ice is 0.030. if janessa releases her stone with an initial velocity of 4.82 m/s, how far will the stone travel before it comes to rest? 16.6 m 34.0 m 47.3 m 39.5 m

Explanation:

Step1: Calculate the acceleration

According to Newton's second law \(F = ma\), and the frictional force \(F_f=\mu N\). On a horizontal surface \(N = mg\), so \(F_f=\mu mg\). Then \(a=\frac{F_f}{m}=\mu g\). Given \(\mu = 0.030\) and \(g = 9.8\ m/s^2\), we have \(a=0.030\times9.8\ m/s^2 = 0.294\ m/s^2\). The acceleration is negative because it is decelerating, \(a=- 0.294\ m/s^2\).

Step2: Use the kinematic equation \(v^2=v_0^2 + 2ax\)

We know that \(v = 0\) (comes to rest), \(v_0=4.82\ m/s\), and \(a=-0.294\ m/s^2\). Rearranging the equation for \(x\) gives \(x=\frac{v^2 - v_0^2}{2a}\). Substitute the values: \(x=\frac{0-(4.82)^2}{2\times(- 0.294)}\).

$$x=\frac{-23.2324}{-0.588}$$

$$x = 39.5\ m$$

Answer:

39.5 m