Work-Energy Theorem EX 8

(a) The work is equal to the area under the force-position graph. From symmetry, we can see that from 0 m to 6 m, the area under the graph below the x-axis is equal to the area under the graph above the x-axis. We only need to calculate the area under the graph from 6 m to 10 m. The work done is

$W=20J+10J=30J$.

(b) The change in kinetic energy is equal to the work done:

$\Delta K=W=30J$

${\frac {1}{2}}m{v_{2}}^{2}-{\frac {1}{2}}m{v_{1}}^{2}=30J$

${\frac {1}{2}}(2){v_{2}}^{2}-{\frac {1}{2}}(2)(2)^{2}=30J$

${\frac {1}{2}}(2){v_{2}}^{2}-4=30J$

${v_{2}}^{2}={34J \over 1kg}$

$v_{2}=5.83m/s$