Question

Part C and D please, part A=291.2 and part B=33.8

Answer 1

Given,

The mass of the skater, m=63 kg

The coefficient of static friction, μs=0.4

The coefficient of the kinetic friction, μk=0.02

F₁=242 N

F₂=162 N

(c) The static friction is given by,

`f_s=N\mu_s`

Where N is the normal force.

The normal force acting on the skater is

`N=mg`

Where g is the acceleration due to gravity.

Therefore the static friction is given by,

`f_s=mg\mu_s`

On substituting the known values,

`\begin{gathered} f_s=63*9.8*0.4 \\ =246.96\text{ N} \end{gathered}`

Therefore the static friction on the skater is 246.96 N

d)The net force acting on the skater is

`\begin{gathered} F_{\text{net}}=ma_{} \\ =F_{\text{tot}}-f \\ =F_{\text{tot}}-N\mu_k \\ =F_{\text{tot}}-mg\mu_k \end{gathered}`

On substituting the known values,

`\begin{gathered} 63a=291.2-63*9.8*0.02 \\ a=(278.85)/(63) \\ =4.43m/s^2 \end{gathered}`

Thus the acceleration of the skater is 4.43 m/s²