A small particle moves to position $5 \hat{i}-2 \hat{j}+\hat{k}$ from its initial position $2 \hat{i}+3 \hat{j}-4 \hat{k}$ under the action of force $5 \hat{i}+2 \hat{j}+7 \hat{k} N$. The value of work done will be ________$J$

Question

A curve is given between the potential energy $U$ of a particle and its position on the $x$-axis as shown. Given: \[ \tan\theta_1 = 1,\qquad \tan\theta_2 = 3,\qquad \tan\theta_3 = -\frac{1}{2} \] If $F_{AB}$ is the force acting on the particle during motion from $A$ to $B$, similarly $F_{BC}$, $F_{CD}$ and $F_{DE}$ are the forces during $B$ to $C$, $C$ to $D$ and $D$ to $E$ respectively, arrange the magnitudes of these forces in decreasing order.

Answer 1

To determine the work done by the force on the particle, we use the formula for work done, which is the dot product of the force vector F and the displacement vector s:
W = F · s

First, calculate the displacement vector s:
The initial position vector is r_i = 2i + 3j − 4k.
The final position vector is r_f = 5i − 2j + k.
Displacement vector s = r_f − r_i
=> s = (5i − 2j + k) − (2i + 3j − 4k)
=> s = (5 − 2)i + (−2 − 3)j + (1 + 4)k
=> s = 3i − 5j + 5k.

The force vector is F = 5i + 2j + 7k N.

Calculate the dot product:
F · s = (5i + 2j + 7k) · (3i − 5j + 5k)
= 5×3 + 2×(−5) + 7×5
= 15 − 10 + 35
= 40 J.

The work done is 40 Joules, which falls within the specified range of 40 to 40 J. Thus, the computed value is consistent with the expected outcome.

A small particle moves to position $5 \hat{i}-2 \hat{j}+\hat{k}$ from its initial position $2 \hat{i}+3 \hat{j}-4 \hat{k}$ under the action of force $5 \hat{i}+2 \hat{j}+7 \hat{k} N$. The value of work done will be ________$J$

Correct Answer: 40

Solution and Explanation

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