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:

$\mathrm{\Delta }K=W=30J$

$\frac{1}{2}m{v_2}^2-\frac{1}{2}m{v_1}^2=30J$

$\frac{1}{2}(2){v_2}^2-\frac{1}{2}(2)(2{)}^2=30J$

$\frac{1}{2}(2){v_2}^2-4=30J$

${v_2}^2=\frac{34J}{1kg}$

$v_2=5.83m/s$