Question

A force F=( 3i + 4j)N acts on a particle moving in xy plane starting from the origin the particle first go along x-axis to the point (4,0)m and then parallel to y-axis to the point (4,3 )m the total work done by the force on the particle

Answer 1

Answer: The total work done is 24 J

Explanation:

Given that,

Force F = (3i+4j) N

Point along x-axis = (4,0)

Point along y-axis = (4,3)

The displacement along x-axis, d = 4i

The displacement along y-axis, d = 3j

Now, The total work done is

$W = F\cdot d_{x}+F\cdot d_{y}$

$W = (3i+4j)\cdot (4i+0j) +(3i+4j)\cdot (0i+3j)$

$W = 12+12$

$W = 24 J$

Hence, the total work done is 24 J

Answer 2

"Answer: Work done = 24 J

Given:

F = (3i + 4j)

Displacement vector on the x-axis = 4i and on y- axis = 3j

Total displacement = 4i + 3j

We know that, Work = dot product of Force and displacement = F.D

=(3i + 4j).(4i + 3j)

$=\quad 12{ i }^{ 2 }\quad +\quad 12{ j }^{ 2 }$

= 12 + 12 = 24 J

Work done = 24 J"