Answer:
(a) v = 1.71 m/s
(b) μ = 0.005
Explanation:
(a)
Using the law of conservation of the momentum:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\[/tex]
where,
m₁ = mass of person = 61.1 kg
m₂ = mass of sled = 16.1 kg
u₁ = initial speed of the person = 2.16 m/s
u₂ = initial speed of the sled = 0 m/s
v₁ = v₂ = v = final speeds of both the person and the sled = ?
Therefore,
[tex](61.1\ kg)(2.16\ m/s)+(16.1\ kg)(0\ m/s)=(61.1\ kg)(v)+(16.1\ kg)(v)\\\\v = \frac{131.976\ kgm/s}{77.2\ kg}[/tex]
v = 1.71 m/s
(b)
The kinetic energy lost by the sled must be equal to the frictional energy:
K.E = fd
[tex]\frac{1}{2}mv^2=\mu Rd = \mu Wd\\\\\frac{1}{2}mv^2=\mu mgd\\\\\frac{1}{2}v^2=\mu g\\\\\mu = \frac{v^2}{2gd}[/tex]
where,
μ = coefficient of kinetic friction = ?
d = distance covered = 30 m
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]\mu = \frac{(1.71\ m/s)^2}{(2)(9.81\ m/s^2)(30\ m)}[/tex]
μ = 0.005
