Question

A particle of mass 5gm is moving with speed 10cm/s along a straight line. A force of 10√2
dyne is applied at angle 45 degree with the initial line of motion of particle. Find change in KE of
particle in 1second.​

Answer 1

Given:

A particle of mass 5gm is moving with speed 10cm/s along a straight line. A force of 10√2

dyne is applied at angle 45 degree with the initial line of motion of particle.

To find:

Change in KE in 1 second.

Diagram:

$\boxed{\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2, 2){\line(0, 1){2}}\put(2, 4){\line(1, 0){2}}\put(4, 4){\line(0, -1){2}}\put(4, 2){\line(-1, 0){2}}\put(3,3){\vector(1,0){2}}\put(1,2){\line(1,0){4}}\put(5,3){u}\put(3,3){\vector(1,1){2}}\put(5.25,5){F}\end{picture}}$

Calculation:

Initial velocity along x axis = 10 cm/s

So , component of acceleration along x axis:

$a_{x} = \dfrac{10 \sqrt{2} }{5} \cos( {45}^{ \circ} )$

$= > a_{x} = \dfrac{10 \sqrt{2} }{5} \times \dfrac{1}{ \sqrt{2} }$

$= > a_{x} = 2 \: cm {s}^{ - 2}$

Initial velocity along y axis = 0 cm/s

So, component of acceleration along y axis:

$a_{y} = \dfrac{10 \sqrt{2} }{5} \sin( {45}^{ \circ} )$

$= > a_{y} = \dfrac{10 \sqrt{2} }{5} \times \dfrac{1}{ \sqrt{2} }$

$= > a_{y} = 2 \: cm {s}^{ - 2}$

So , Initial KE:

$= \dfrac{1}{2} m {u}^{2}$

$= \dfrac{1}{2} \times 5 \times {(10)}^{2}$

$= 250 \: g \: cm \: {s}^{ - 2}$

So , final KE:

$= \dfrac{1}{2} m {(v_{x})}^{2} + \dfrac{1}{2} m {(v_{y})}^{2}$

$= \dfrac{1}{2}m \bigg \{ {(v_{x})}^{2} + {(v_{y})}^{2} \bigg \}$

$= \dfrac{1}{2}m \bigg \{ {(10 + 2 \times 1)}^{2} + {(0 + 2 \times 1)}^{2} \bigg \}$

$= \dfrac{1}{2}m \bigg \{ {(12)}^{2} + {(2)}^{2} \bigg \}$

$= \dfrac{1}{2}m \bigg \{ 144 + 4 \bigg \}$

$= \dfrac{1}{2} \times 5 \times 148$

$= 370 \: g \: cm \: {s}^{ - 2}$

So, change in KE

$\Delta KE = 370 - 250 = 120 \: gcm {s}^{ - 2}$

So, final answer is:

$\boxed{ \bf{ \large{\Delta KE = 120 \: gcm {s}^{ - 2} }}}$