Force–Position Graphs, Dissipated Energy and Gravitational Potential Energy at a Distance

Work is the area under a force–position graph

$W=Fd\cos\theta$ only works for a constant force. When the force varies with position, the work done is the area between the $F$–$x$ curve and the axis, with area below the axis counted negative. This is where the spring formula comes from: the graph of $F=kx$ is a straight line through the origin, and the triangle under it from $0$ to $x$ has area $\tfrac12(x)(kx)=\tfrac12kx^2$. A force that rises linearly from $0$ to $40$ N over $0{,}5$ m does $\tfrac12(0{,}5)(40)=10$ J of work, not $20$ J.

Where the energy goes with friction

Kinetic friction converts mechanical energy into thermal energy in exactly the amount

where $d$ is the distance slid, not the displacement. On a round trip the displacement is zero but friction still removes energy on both legs. The full energy statement is $K_i+U_i+W_{\text{ext}}=K_f+U_f+\Delta E_{\text{thermal}}$; on AP free-response questions you will be asked to name where the missing energy went, so say "thermal energy of the block and surface", not "lost".

Gravitational potential energy far from Earth

$U_g=mgh$ assumes $g$ is constant, which is only true near the surface. The general expression, with the zero chosen at infinite separation, is

It is negative everywhere and increases toward zero as $r$ grows — moving away from a planet always raises potential energy, exactly as climbing does. The negative sign is a bookkeeping choice, not a sign that energy is "missing". The formula gives the escape speed directly: total energy zero at infinity means $\tfrac12mv^2-\frac{GMm}{R}=0$, so $v_{\text{esc}}=\sqrt{2GM/R}$, about $11{,}2$ km/s for Earth, independent of the mass of the object.