block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8,−6)N→F3= 14N[↑].The normal force and the weight force act perpendicular to the plane (along thez-axis) andcancel everywhere and always.At the instantt= 0sthe block is at rest at the origin,~r0= (0,0)m.Employing dynamics methods we shall determine the speed of the block at the timet= 2s,when it reaches the point~r2= (8,8)m.(a) (i) Draw and label each of the forces on the axes provided below.(ii) Determine the net force on the block.

The force F1 acting at 45 degrees would have multiplication factors of [tex]\frac{\sqrt{2} }{2}[/tex] on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is [tex]F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N[/tex]

y-component of F1 is [tex]F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N[/tex]

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = [tex]4\,\,\,m/s^2[/tex]

y-component of acceleration: 20 N / 5 kg = [tex]4\,\,\,m/s^2[/tex]

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: [tex]v_x=4\,*\,2=8\,m/s[/tex]

y-component of the velocity is: [tex]v_y=4\,*\,2=8\,m/s[/tex]