Question

. A particle moving in the xy plane undergoes a displacement (2.0i+3.0j) m as a constant force F (5.0i+ 2.0) acts on the particle. a) Calculate the magnitudes of the displacement and the force b) Calculate the work done by c) Calculate the angle between the force and displacement.

Answer 1

Displacement, d=$2i+3j$

Force, F= $5i+2j$

(a) Magnitude of Displacement, $|d|=\sqrt{(2)^2+(3)^2}=\sqrt{4+9}=\sqrt{13}$

Magnitude of force, $|F|=\sqrt{(5)^2+(2)^2}=\sqrt{25+4}=\sqrt{29}$

(b) Work done, $W=F.d=(2i+3j).(5i+2j)=10+6=16 joule$

(c) As we know $W=|F||d| cos\theta$

$16=\sqrt{13}.\sqrt{29} cos\theta$

$cos\theta=\dfrac{16}{\sqrt{13}.\sqrt{29}}=0.824$

$\theta=cos^{-1}(0.824)=34.5^{\circ}$