A bar magnet is pushed into a closed coil of resistance 4.0 Ω at a steady speed…

Step 1. The heat. $H = I^2Rt = (0.15\ \text{A})^2 \times 4.0\ \Omega \times 0.30\ \text{s} = 0.027\ \text{J}$, that is $27$ mJ.

Step 2. The work. The magnet moves at a steady speed, so its kinetic energy does not change. The energy in the coil must come from the work of the hand. So $W = 27$ mJ at least.

Step 3. The force. $W = Fd$, so $F = \dfrac{W}{d} = \dfrac{0.027\ \text{J}}{0.060\ \text{m}} = 0.45$ N.

Step 4. The open coil. An open coil has an emf but no current. There is no induced field and no force, so the hand needs no extra force and does no extra work.