Figure 5.8 shows (x, t), (y, t ) diagram of a particle moving in 2-dimensions If the particle has a mass of 500 g, find the force (direction and magnitude) acting on the particle.

This is a Short Answer type Questions as classified in NCERT Exemplar

Explanation- from the diagram x= t=dx/dt=1m/s

So a_x= 0

From the diagram y =t²

dy/dt=2t  or a_y=d²y/dt²=2m/s²

I_y= ma_y=500 (10^-3) (2) =1N

F_x=ma_x=0

F= $\sqrt[]{F_x^2+F_y^2}$ = 1N

<div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"></picture>This is a Short Answer type Questions as classified in NCERT Exemplar</div><div class="figure"><picture><img src="https://images.shiksha.com/mediadata/images/1751023513phpjNo3q0.jpeg" alt="" width="559" height="307" loading="lazy"></picture></div></div><p><strong><span lang="EN-IN">Explanation- </span></strong><span lang="EN-IN">from the diagram x= t=dx/dt=1m/s</span></p><p><span lang="EN-IN">So a_x= 0</span></p><p><span lang="EN-IN">From the diagram y =t²</span></p><p><span lang="EN-IN">dy/dt=2t&nbsp; or a_y=d²y/dt²=2m/s²</span></p><p><span lang="EN-IN">I_y= ma_y=500 (10^-3) (2) =1N</span></p><p><span lang="EN-IN">F_x=ma_x=0</span></p><p><span lang="EN-IN">F=<span class="mathml" contenteditable="false"> <math> <mroot> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></span><span lang="EN-IN">= 1N</span></p>

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