Final answer:
The speed of the block at point B is 6.4 m/s. The work of friction along the horizontal surface is 0.8576 J. The coefficient of kinetic friction along the horizontal surface is 0.413.
Step-by-step explanation:
To determine the speed at point B, we can use the principle of conservation of mechanical energy. We know that the block starts at rest at point A and slides to point B. At point B, where the speed of the block is given as 6.4 m/s, the only form of energy it has is kinetic energy. Therefore, the mechanical energy at point B is equal to the kinetic energy at point B:
KE = 0.5mvb2
where m is the mass of the block and vb is the speed at point B. Plugging in the given values:
KE = 0.5(0.210 kg)(6.4 m/s)2
Solving for KE gives us KE = 0.8576 J.
Therefore, the speed of the block at point B is 6.4 m/s.
For part (b) of the question, we can use the work-energy theorem to find the work of friction along the horizontal surface. The work done by friction is equal to the initial mechanical energy minus the final mechanical energy:
- The initial mechanical energy is given by the kinetic energy at point B, which we calculated to be 0.8576 J.
- The final mechanical energy is zero, as the block comes to rest at point C.
Therefore, the work of friction along the horizontal surface is 0.8576 J.
To find the coefficient of kinetic friction, we can use the equation:
W = μkFn
where W is the work of friction, μk is the coefficient of kinetic friction, and Fn is the normal force. Plugging in the given values:
0.8576 J = μk(0.210 kg)(9.8 m/s^2)
Solving for μk gives us μk = 0.413.