Question

Question:A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the xy plane. The $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$puck has a velocity of $\mathbf{3}\mathbf{.}\mathbf{00}\hat{\mathbf{i}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at one instant. Eight seconds later, its velocity is $\left(8.00\hat{i}+10.00\hat{j}\right)\mathbf{)}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude.

Short answer

(a) The components of force are $2.5\stackrel{⏜}{\mathrm{i}}\mathrm{N}$and $5.0\stackrel{⏜}{\mathrm{j}}\mathrm{N}$.

(b) The magnitude of force is $5.6\mathrm{N}$.

Step-by-step solution

A concept and Identification of given data

The acceleration of the object is the variation in the velocity of the object with position and time.

Consider the given data as below.

The mass of puck,$m=4\mathrm{kg}$

The initial velocity of puck, ${\mathrm{u}}_{\mathrm{x}}=3.00\hat{\mathrm{i}}\mathrm{m}/\mathrm{s}$

The final velocity of puck, ${\mathrm{v}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{v}}_{\mathrm{y}}\hat{\mathrm{j}}=\left(8.00\hat{\mathrm{i}}+10.00\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}$

The duration for motion,$t=8s$

(a) Determination of resultant force acting on object

The horizontal component of the force is given as

$F_x=m\left(\frac{v_x-u_x}{t}\right)$

Substitute all the values in the above equation.

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{x}}=4\mathrm{kg}\times \left(\frac{\left(8\hat{\mathrm{i}}-3\hat{\mathrm{i}}\right)\mathrm{m}/\mathrm{s}}{8\mathrm{s}}\right) \\ {\overrightarrow{\mathrm{F}}}_{\mathrm{x}}=0.5\times 5\hat{\mathrm{i}}\mathrm{N} \\ {\overrightarrow{\mathrm{F}}}_{\mathrm{x}}=2.5\hat{\mathrm{i}}\mathrm{N} \end{gathered}$$

The vertical component of the force is given as

$F_y=m\left(\frac{v_y-u_y}{t}\right)$

Substitute all the values in the above equation.

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_y=4\mathrm{kg}\times \left(\frac{\left(10\hat{\mathrm{j}}-0\hat{\mathrm{j}}\right)\mathrm{m}/\mathrm{s}}{8\mathrm{s}}\right) \\ {\overrightarrow{\mathrm{F}}}_{\mathrm{y}}=0.5\times 10\hat{\mathrm{j}}\mathrm{N} \\ {\overrightarrow{\mathrm{F}}}_{\mathrm{y}}=5\hat{\mathrm{j}}\mathrm{N} \end{gathered}$$

Therefore, the components of force are $2.5\stackrel{⏜}{\mathrm{i}}\mathrm{N}$and $5.0\hat{\mathrm{j}}\hspace{0.17em}\mathrm{N}$.

(b) Determination of magnitude of force

The magnitude of force is given as

$F=\sqrt{{\left(F_x\right)}^2+{\left(F_y\right)}^2}$

Substitute all the values in the above equation.

$\begin{array}{rcl}F& =& \sqrt{{\left(2.5\hat{\mathrm{i}}\mathrm{N}\right)}^2+{\left(5.0\hat{\mathrm{j}}\mathrm{N}\right)}^2}\\ & =& \sqrt{6.25+25}\mathrm{N}\\ & =& \sqrt{31.25}\mathrm{N}\\ & =& 5.6\mathrm{N}\end{array}$

Hence, the magnitude of force is $\begin{array}{rcl}& & 5.6\mathrm{N}\end{array}$.