A 5 kg box is along a horizontal force. Three horizontal forces act on the box,

Question

A 5 kg box is along a horizontal force. Three horizontal forces act on the box, Force F_1 has a magnitude of 15N and is directed rightward, force F_2 has a magnitude of 15 N and is directed left force F_3has a magnitude of 5 N and is directed leftward. At t = 0, the velocity of the blockis5m/s to the right. Let rightward be the positive direction. is the acceleration of the box positive, negative or zero ? What is the magnitude of the box's acceleration? What is the blocks displacement from s to r = 3 s?

Explanation / Answer

Net force acting on the body = (15-15-5)N=-5N Acceleration due to net force =(net force/mass)=-5/5=-1 kg/m^2 At t=3s the displacement(s) is given by S s=ut+0.5*a*t^2; u=initial velocity s=(5*3-0.5*1*9 ) m s =10.5m As the net force acting on leftward which is in -ve direction .so the displacement would be -10.5m

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