Question

this is a practice question not a graded assingment.The friction force is 30 N

Answer 1

First, let's calculate the horizontal component of the applied force:

$\begin{gathered} F_x=F\cdot\cos35°\\ \\ F_x=70\cdot0.819\\ \\ F_x=57.33\text{ N} \end{gathered}$

Now, calculating the works, we have:

a.

To calculate this work, we use the horizontal component of the applied force:

$\begin{gathered} W_=F_x\cdot d\\ \\ W=57.33\cdot10\\ \\ W=573.3\text{ J} \end{gathered}$

b.

To calculate this work, we use the friction force only:

$\begin{gathered} W_f=F_f\cdot d\\ \\ W_f=-30\cdot10\\ \\ W_f=-300\text{ J} \end{gathered}$

(We use a negative force because it is against the movement)

c.

The total work on the wagon is given by:

$\begin{gathered} W_{net}=W+W_f\\ \\ W_{net}=573.3-300\\ \\ W_{net}=273.3\text{ J} \end{gathered}$