A force of 7N, making an angle with the horizontal, acting on an object displaces it by 0.5m along the horizontal direction. If the object gains K.E of 2J, what is the horizontal component of the force?
A) 2
B) 4
C) 1
D) 14

Hint: Apply work energy theorem which states that net work done by the forces on an object equals change in its kinetic energy. The displacement takes place in horizontal direction therefore only horizontal components of force will perform work.

Complete answer:
As question says force of 7N be acting at the angle with horizontal direction. Since displacement is in horizontal direction, only horizontal components of force will perform work.
First we have to write given values
d =0.5m is the object displacement in the direction of horizontal.
$\Delta K$=2J is the change in kinetic energy.
F =7N is the force applied on the object.
${{\text{F}}_{\text{H}}}$ is the force acting in the direction of horizontally where the work is done. In this question we have to find its value.
We know that work energy theorem which states that net work done by the forces on an object equals change in its kinetic energy, write it in mathematical form.
${\text{W = }}\Delta {\text{K}}$
W is the work done which takes place in horizontal direction so, we can write
${F_H} \times 0.5 = 2J$
$ \Rightarrow {F_H} = \dfrac{2}{{0.5}}$
$\therefore {F_H} = 4N$

Hence the correct answer is 4N. which is given in option B.

Additional information:
Force is nothing but is push or pull. The S.I unit of force is newton represented by N.
Acceleration means increase in the velocity of an object.
Memorize the three laws of motion given by Newton. These three laws of motion relate to motion of an object when force is acting on it.

Note:
If force applied on an object then there is change in its motion, direction, speed and also its shape. Remember there is a very small difference between the force applied and the pressure applied on an object.so, the pressure is always given as the force per unit area.