Question

(1) Two external forces,(40,70,0)Nand (20,10,0)N, act on a system. What is the net force acting on the system? (2) A hockey puck initially has momentum (0,2,0)kg.m/s. It slides along the ice, gradually slowing down, until it comes to a stop. (a) What was the impulse applied by the ice and the air to the hockey puck? (b) It took 3 seconds for the puck to come to a stop. During this time interval, what was the net force on the puck by the ice and the air (assuming that this force was constant)?

Short answer

1. The net force on the system is (60,-60)N.
2. (a)The ice's impulse acting on the hockey puck is $∆\overrightarrow{\mathrm{p}}=\left(0,2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}$.

(b)The net force on the hockey puck is (0,-0.667,0)N.

Step-by-step solution

Understanding the net force and its formula

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,

${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\overrightarrow{{\mathbf{F}}_{\mathbf{1}}}\mathbf{+}\overrightarrow{{\mathbf{F}}_{\mathbf{2}}}$

Here,$\overrightarrow{{\mathrm{F}}_1}$is the first external force and $\overrightarrow{{\mathrm{F}}_1}$is the second external force.

(1) Finding the net force on the system

Let insert (40,-70,0)N for ${\overrightarrow{\mathrm{F}}}_1$and (20,10,0)N for$\overrightarrow{F_2}$ into the formula of the net force.

$$\begin{gathered} \overrightarrow{{\mathrm{F}}_{\mathrm{net}}}=\left(40,-70,0\right)\mathrm{N}+\left(20,10,0\right)\mathrm{N} \\ =\left(60,-60,0\right)\mathrm{N} \end{gathered}$$

As a result, the net force on the system is (60,-60,0)N .

2 (a) Find impulse on the hockey puck

The ice's impulse on the hockey puck is,

$∆\overrightarrow{p}={\overrightarrow{p}}_f-\overrightarrow{p_i}$

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

Substitute (0,0,0)kg.m/s for $\overrightarrow{p_f}$and (0,2,0)kg.m/s for$\overrightarrow{p_i}$ into the ice’s impulse.

$$\begin{gathered} ∆\overrightarrow{\mathrm{p}}=\left(0,0,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}-\left(0,2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =\left(0,-2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the ice's impulse acting on the hockey puck is (0,2,0 kg.m/s.

2 (b) Determine net force on the hockey puck

The ice's impulse on the hockey puck is,

$∆\overrightarrow{p}=\overrightarrow{F_{Net}}∆t.$

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

Now rearrange the equation for $\overrightarrow{F_{Net}}$.

${\overrightarrow{F}}_{Net}=\frac{∆\overrightarrow{p}}{∆t}$

Substitute (0,-2,0)kg.m/s for $∆\overrightarrow{p}$and 3.0 s for$∆t$ into the above formula.

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{Net}}=\frac{\left(0,2,0\right)\mathrm{kg}.\mathrm{m}/\mathrm{s}}{3.0\mathrm{s}}.\left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/\mathrm{s}}\right) \\ =\left(0,-0.667,0\right)\mathrm{N} \end{gathered}$$

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