91Ó°ÊÓ

The vector addition of all available forces acting on any particular object is the net force.

The system's overall net force is given as,

$\underset{F_{net}}{→}=\underset{F_1}{→}+\underset{F_2}{→}$

Here, $\underset{F_1}{→}$is the first external force and $\underset{F_2}{→}$is the second external force.

Let insert $\left\{40,-70,0\right\}N$for $\underset{F_1}{→}$and $\left\{20,10,0\right\}$N for $\underset{F_2}{→}$into the formula of net force.

$\underset{F_{net}}{→}$= $\left\{40,-70,0\right\}$N $+\left\{20,10,0\right\}$N = $\left\{60,-60,0\right\}$N

As a result, the net force on the system is$\left\{60,-60,0\right\}N$

The ice's impulse on the hockey puck is,

$∆\underset{p}{→}=\underset{p_f}{→}-\underset{p_j}{→}$

The end momentum of the hockey puck is $\underset{p_f}{→}$while the starting momentum of the hockey puck is$\underset{p_i}{→}$

Substitute $\left\{0,0,0\right\}$$kg.m/s$for $\underset{p_f}{→}$and $\left\{0,2,0\right\}$$kg.m/s$for $\underset{p_i}{→}$into the ice’s impulse.

$∆\underset{p}{→}=\left\{0,0,0\right\}kg.m/s-\left\{0,2,0\right\}kg.m/s=\left\{0,-2,0\right\}kg.m/s$

Thus, the ice's impulse acting on the hockey puck is$\left\{0,-2,0\right\}kg.m/s$

The ice's impulse on the hockey puck is,

$∆\underset{p}{→}=\underset{F_{ne}}{→}∆t$

The end momentum of the hockey puck is $\underset{p_f}{→}$, the net force operating on the hockey puck is $\underset{F_{Net}}{→}$, the time is $∆t$, and the initial momentum of the hockey puck is$\underset{p_i}{→}$

Now rearrange the equation for$\underset{F_{Net}}{→}$

$\underset{F_{Net}}{→}=\frac{∆\underset{p}{→}}{∆t}$

Substitute $\left\{0,-2,0\right\}$$kg.m/s$for $∆\underset{p}{→}$and $3.0s$for $∆t$into the above formula.

$\underset{F_{Net}}{→}$$=\frac{\left\{0,-2,0\right\}kg.m/s}{3.0s}\left(\frac{1N}{1kg.m/s^2}\right)=\left\{0,-0.667,0\right\}$

Thus, the net force on the hockey puck is$\left\{0,-0.667,0\right\}N$