Question
A force $6 \hat{i}+3 \hat{j}+\hat{k}$ Newton displaces a particles from $A(0,3,2)$ to $B(5,1,6)$. Find the work done.(A) $10 \mathrm{~J}$(B) $22 \mathrm{~J}$(C) $32 \mathrm{~J}$(D) $\underline{41 J}$
Step 1
The initial position vector $\vec{S_1}$ is given by the coordinates of point A and the final position vector $\vec{S_2}$ is given by the coordinates of point B. So, we have: \[\vec{S_1} = 0\hat{i} + 3\hat{j} + 2\hat{k}\] \[\vec{S_2} = 5\hat{i} + 1\hat{j} + Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A uniform force of $(3 \hat{i}+\hat{j})$ newton acts on a particle of mass $2 \mathrm{~kg}$. Hence the particle is displaced from position $(2 \hat{i}+\hat{k}) \mathrm{m}$ to position $(4 \hat{i}+3 \hat{j}-\hat{k}) \mathrm{m}$. The work done by the force on the particle is: (a) $9 \mathrm{~J}$ (b) $6 \mathrm{~J}$ (c) $13 \mathrm{~J}$ (d) $15 \mathrm{~J}$
WORK, ENERGY AND POWER
Work, Energy, and Power
$\Lambda$ particle in cquilibrium displaces from $(1,1,2) \mathrm{m}$ to $(2,0,3) \mathrm{m} .$ If onc of the forces acting on the particle is $(\hat{i} \quad \hat{j} \mid 3 \hat{k}) \mathrm{N}$, total work done by the remaining forces acting on the particle is (a) $5 \mathrm{~J}$ (b) $-5 \mathrm{~J}$ (c) Zero (d) $10 \mathrm{~J}$
Work, Energy, Power and Circular Motion
Section B
A particle is acted upon by a conservative force $F=(7 \hat{i}-6 \hat{j}) \mathrm{N}$. The work done by the force when the particle moves from origin $(0,0)$ to the position $(-3 \mathrm{~m}$, $4 \mathrm{~m}$ ) is given by (A) $3 \mathrm{~J}$ (B) $10 \mathrm{~J}$ (C) $-45 \mathrm{~J}$ (D) None of these
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD