The only force acting on a 2 kg body that is moving in xy- plane has a magnitude of 5 N . The body initially has a velocity of 4 m/s in the positive x -direction. Some time later, the body has a velocity of 6 m/s in the positive y- direction. The work done on the body by the 5 N force during this time is

Question

The only force acting on a 2 kg body that is moving in xy- plane has a magnitude of 5 N . The body initially has a velocity of 4 m/s in the positive x -direction. Some time later, the body has a velocity of 6 m/s in the positive y- direction. The work done on the body by the 5 N force during this time is (A) 20 J (B) 40 J (C) 50 J (D) 72 J

Tardigrade Admin · Accepted Answer

Given, m = 22 kg , V1=4 m /s and V2=6 m /s F=5 N According to work-energy theorem, ∫ Fnet dx=Δ zkE=(KE)2-(KE)1 =(1/2)mv22-(1/2)mv12 =[(6)2-(4)2](2/2) [putting values] =[36-16]=20 J

20 J

40 J

50 J

72 J

Given, m=22kg , V1​=4m/s and V2​=6m/s F=5N According to work-energy theorem, ∫Fnet​dx=ΔzkE=(KE)2​−(KE)1​ =21​mv22​−21​mv12​ =[(6)2−(4)2]22​ [putting values] =[36−16]=20J

Given, $m = 22 k g$ ,
$V_{1} = 4 m / s$ and $V_{2} = 6 m / s$ $F = 5 N$
According to work-energy theorem,
$\int F_{n e t} d x = Δ z k E = (K E)_{2} - (K E)_{1}$
$= \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2}$
$= [(6)^{2} - (4)^{2}] \frac{2}{2}$ [putting values]
$= [36 - 16] = 20 J$