Question

An object of mass 2kg is released at origin of x-y plane, where its potential energy changes as 
U = (2x – 4y) joules. Find speed of object after 3 sec of release. (use
sqrt 5  = 2.24
​

Answer 1

Given:

An object of mass 2kg is released at origin of x-y plane, where its potential energy changes as U = (2x – 4y) joules.

To find:

Speed of object after 3 seconds of release.

Calculation:

First of all, let's find out the force in the X and Y direction:

$\therefore \: F_{x} = - \dfrac{ \partial U}{ \partial x}$

$\implies\: F_{x} = - \dfrac{ \partial (2x - 4y)}{ \partial x}$

$\implies\: F_{x} = - 2 \: N$

$\implies\: a_{x} = - \dfrac{2}{2} = 1 \: m {s}^{ - 2}$

--------------------------------------------------------------

$\therefore \: F_{y} = - \dfrac{ \partial U}{ \partial y}$

$\implies\: F_{y} = - \dfrac{ \partial (2x - 4y)}{ \partial y}$

$\implies\: F_{y} = 4 \: N$

$\implies\: a_{y} = \dfrac{4}{2} = 2 \: m {s}^{ - 2}$

---------------------------------------------------------------

So, X axis velocity of object after 3 seconds.

$\therefore \: v_{x} = 0 + a_{x}t$

$\implies \: v_{x} = ( - 2) \times 3$

$\implies \: v_{x} = - 6 \: m {s}^{ - 1}$

So , Y axis velocity of object after 3 seconds.

$\therefore \: v_{y} = 0 + a_{y}t$

$\implies \: v_{y} = ( 4) \times 3$

$\implies \: v_{y} = 12 \: m {s}^{ - 1}$

So, net velocity will be:

$\therefore \: v_{net} = \sqrt{ {(v_{x})}^{2} + {(v_{y})}^{2} }$

$\implies \: v_{net} = \sqrt{ {( - 6)}^{2} + {(12)}^{2} }$

$\implies \: v_{net} = \sqrt{ 36+ 144}$

$\implies \: v_{net} = \sqrt{180}$

$\implies \: v_{net} = \sqrt{5 \times 36}$

$\implies \: v_{net} = 6\sqrt{5}$

$\implies \: v_{net} = 6 \times 2.24$

$\implies \: v_{net} = 13.44 \: m {s}^{ - 1}$

So, net velocity after 3 seconds is 13.44 m/s.

Answer 2

Answer:

Ans. 6.72

Explanation:

. U = 2x – 4y

Fx

=

U

2

x



  



. Fy

=

U

4

y



 



F 2i 4j ˆ ˆ   



ˆ ˆ a i 2j   



v u at  

  

  u 0 i 2j 3 ˆ ˆ     



ˆ ˆ v 3i 6j   



u 9 36 3 5 3 2.24      = 6.72