Work-Energy Theorem EX 5

(a) We choose the x-axis pointing to the right in the direction of the displacement. The x-component of the displacement is $\mathrm{\Delta }x=10m$ in this coordinate system. Then we draw a free body diagram which includes these forces: $F_G$, $F_N$, $F_R$ and the force exerted by the man $F$ - see the diagram below:

Now we are ready to compute the work done by each force. The x-component of $F_G$, as well as the x-component of the normal force $F_N$, are zero and therefore the work done by these forces is equal to zero. Remember that whenever a force is perpendicular to the displacement the work done by this force is always zero.

The x-component of the force exerted by the man is $F_x=80\cos (-20^{\circ })=75.2N$. The work done by this force is $W_F=(75.2)(10)=752J$.

The x-component of the resistive force exerted on the lawnmower is ${F_R}_x=10\cos (180^{\circ })=-10N$. The work done by this force is $W_{F_R}=(-10)(10)=-100J$.

The total work done on the lawnmower is

$W_{total}=752+(-100)=652J$

(b) The sum of the y-components of forces is equal to zero. The sum of the x-components is equal $F_{net}=75.2-10=65.2N$. The work done by the net force is

$W_{total}=(65.2)(10)=652J$

We note that both methods yield the same result.