Question

1. a 49 kg ice skater is pushed across the ice in a skating rink by a friend. the coefficient of friction between the skaters ice skates and the ice is 0.08. what force must the friend apply to push the skater at a constant speed across the rink?

Explanation:

Step1: Calculate the normal force

The normal force \(N\) on the skater is equal to the skater's weight. Using the formula \(N = mg\), where \(m = 49\ \text{kg}\) and \(g=9.8\ \text{m/s}^2\).

$$N=49\times9.8 = 480.2\ \text{N}$$

Step2: Calculate the frictional force

The frictional force \(f=\mu N\), where \(\mu = 0.08\) (coefficient of friction) and \(N = 480.2\ \text{N}\) (normal force).

$$f=0.08\times480.2=38.416\ \text{N}$$

Since the skater is moving at a constant speed, the applied force \(F\) must be equal to the frictional force. So \(F = f=38.416\ \text{N}\)

Answer:

$38.416\ \text{N}$