Question

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.

Answer 1

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude:
`√(800) \,\,\,N`

Step-by-step explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of
`(√(2) )/(2)` on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is
`F1_x=12\,√(2) (√(2) )/(2) =12\,\,N`

y-component of F1 is
`F1_y=12\,√(2) (√(2) )/(2) =12\,\,N`

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 =
`4\,\,\,m/s^2`

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

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:
`v_x=4\,*\,2=8\,m/s`

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