Answer
(i) The total mechanical energy is conserved.
(ii) The block's kinetic energy increases as it reaches B.
(iii) The total mechanical energy is not conserved.
a) $KE_B=57.09J$
b) $W_{f_k}=-57.09J$
Work Step by Step
(i) The total mechanical energy of the block is conserved from A to B, because the surface of the slide is frictionless, so all the forces affecting the block's motion are conservative.
(ii) We know that $E=KE+PE$ and $PE=mgh$, meaning for the same object, sitting at a higher position means having a higher PE.
From A to B, because the energies are conserved, $E_A=E_B$
We notice that $A$ is higher than $B$, meaning $PE_A\gt PE_B$; because the energies are conserved, it follows that $KE_A\lt KE_B$. Therefore, the block's kinetic energy increases as it reaches B.
(iii) The total mechanical energy of the block is not conserved from B to C, because kinetic frictional force affects the block's motion, and kinetic frictional force is not conservative.
a) According to the principle of energy conservation, $$E_B-E_A=0$$ $$(KE_B-KE_A)+(PE_B-PE_A)=0$$
We have $PE_B-PE_A=mg(h_B-h_A)=0.41\times9.8(7-12)=-20.09J$
Therefore, $$KE_B-KE_A-20.09=0$$ $$KE_B=20.09+37=57.09J$$
b) From B to C, kinetic friction $f_k$ opposes the motion, reducing the block's speed to 0, meaning $KE_C=0$. Therefore, $$W_{f_k}=KE_C-KE_B=-57.09J$$
