A force of magnitude 50, N acting along hat{i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector along the direction $\hat{i} + \hat{j} + \hat{k}$.Unit vector $

Vedclass · Accepted Answer

$-50 \sqrt{3}\, J$

Vedclass · Answer

(C) The force vector $\vec{F}$ is given by its magnitude multiplied by the unit vector along the direction $\hat{i} + \hat{j} + \hat{k}$.Unit vector $\hat{n} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$.So,$\vec{F} = 50 	imes \frac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}} = \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})\, N$.The displacement vector $\vec{d} = \vec{r}_2 - \vec{r}_1 = (4-5)\hat{i} + (8-9)\hat{j} + (6-7)\hat{k} = -\hat{i} - \hat{j} - \hat{k}\, m$.The work done $W = \vec{F} \cdot \vec{d} = \left[ \frac{50}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) ight] \cdot (-\hat{i} - \hat{j} - \hat{k})$.$W = \frac{50}{\sqrt{3}} (-1 - 1 - 1) = \frac{50}{\sqrt{3}} (-3) = -50\sqrt{3}\, J$.

$A$ force of magnitude $50\, N$ acting along $\hat{i} + \hat{j} + \hat{k}$ displaces a point mass from $(5, 9, 7)$ to $(4, 8, 6)$. The work done by this force during this displacement is:

Similar Questions

Given that a force $\vec{F}$ acts on a body for time $t_1$ and displaces the body by $\vec{d}$. In which of the following cases must the velocity of the body increase?

Two constant forces ${F_1} = 2\hat i - 3\hat j + 3\hat k$ $(N)$ and ${F_2} = \hat i + \hat j - 2\hat k$ $(N)$ act on a body and displace it from the position ${r_1} = \hat i + 2\hat j - 2\hat k$ $(m)$ to the position ${r_2} = 7\hat i + 10\hat j + 5\hat k$ $(m)$. What is the work done in $J$?

$A$ body of mass $8 \ kg$,under the action of a force,is displaced according to the equation $s = \frac{t^2}{4} \ m$,where '$t$' is the time. Find the work done by the force in the first $4 \ s$. (in $J$)

$A$ block of mass $m$ is kept in an elevator which starts moving downward with an acceleration $a$ as shown in the figure. The block is observed by two observers $A$ and $B$ for a time interval $t_0$. The observer $B$ (who is on the ground) finds that the work done by gravity on the block is:

$A$ chain of mass $4 \ kg$ and length $2 \ m$ is hanging such that one-fourth of its length is hanging over the edge of a table. The work done to pull the hanging part onto the table is . . . . . . $J$. (Take $g = 10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module