Question

The work done in moving a particle from a point
(1, 1) to (2, 3) in a plane in force field with potential
U = k(x + y) is
(1) 3k
(2) -3K
(3) K
(4) Zero
​

Answer 1

Answer:  -3K

Explanation:

Answer 2

Answer:

The work done in moving a particle from a point (1, 1) to (2, 3) in a plane in a force field with potential U = k(x + y) is $-3\hat k$ i.e. option (2).

Explanation:

The work done is defined as the dot product of force and displacement between two-point i.e.

$\overrightarrow W=\overrightarrow F.\overrightarrow r$            (1)

The values given in the question are;

$\overrightarrow r_1=(\hat i+\hat j)$

$\overrightarrow r_2=(2\hat i+3\hat j)$

$U = k(x + y)$

$\overrightarrow{F}=-\frac{\delta U}{\delta x} \hat i-\frac{\delta U}{\delta y}\hat j$         (2)

As per question potential energy is,

$\overrightarrow{U}= kx\hat i + ky\hat j$              (3)

By putting equation (3) in equation (2) we get;

$\overrightarrow F= -k\hat i-k\hat j$           (4)

Now we only need to find the displacement vector which can be done as follows;

$\overrightarrow r=\overrightarrow r_2 -\overrightarrow r_1$

$\overrightarrow r=(2\hat i+3\hat j)-(\hat i+\hat j)$

$\overrightarrow r=\hat i+\hat 2j$            (5)

By substituting equations (4) and (5) in equation (1) we get;

$\overrightarrow W=( -k\hat i-k\hat j).(\hat i+\hat 2j)$

$\overrightarrow W=-3\hat k$

Hence, The work done in moving a particle from a point (1, 1) to (2, 3) in a plane in a force field with potential U = k(x + y) is $-3\hat k$ i.e. option (2).