A particle moves under the action of a force $\to F = 20^i + 15^j$ along a straight line 3y+ax=5, where $a$ is a constant. If the work done by the force F is zero, then the value of $a$ is
$\frac{4}{9}$
$\frac{9}{4}$
3
4

Question

A particle moves under the action of a force $\to F = 20^i + 15^j$ along a straight line 3y+ax=5, where $a$ is a constant. If the work done by the force F is zero, then the value of $a$ is
$\frac{4}{9}$
$\frac{9}{4}$
3
4

Toppr Admin · Accepted Answer

For the work done to be 0- the particle should move in perpendicular direction to the application of force-xAF-F-20-i-15-jslope of F-xA0- sF-1520slope of the line is sl-x2212-a3if two lines are perpendicular product of their slopes is -x2212-1hence- slsF-x2212-1-x27F9-slsF-x2212-a-xD7-154-xD7-20-x2212-1-x2234-a-4

A particle moves under the action of a force →F=20^i+15^j along a straight line 3y+ax=5, where a is a constant. If the work done by the force F is zero, then the value of a is499434

For the work done to be 0, the particle should move in perpendicular direction to the application of force.¯F=20^i+15^jslope of F sF=1520slope of the line is sl=−a3if two lines are perpendicular product of their slopes is −1hence, slsF=−1⟹slsF=−a×154×20=−1∴a=4

A particle moves under the action of a force $\to F = 20^i + 15^j$ along a straight line 3y+ax=5, where $a$ is a constant. If the work done by the force F is zero, then the value of $a$ is
$\frac{4}{9}$
$\frac{9}{4}$
3
4