Nuzhat at 16:16, 14 April 2011

2011-04-14T16:16:03Z

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''Helena Dedic''

There are three forces acting on the block: '''<math>F_G</math>''', '''<math>F_N</math>''', '''<math>F</math>'''. 

We can draw the diagram and the free body diagram. 

[[image: N_APP_EX_15_SOLN.png|TOP]] 

Then compute the components of the three forces.

{| border="1" cellpadding="2" style="text-align:center"
!width="50"|Forces
!x-comp
!y-comp
|-

| '''<math>F_G</math>''' || 10 cos(- 127°) = - 6 N || 10 sin(- 127°) = - 8 N

|-

| '''<math>F_N</math>''' || 0 || <math>F_N</math>

|-

| '''<math>F</math>''' || 5 cos(- 37°) = 4 N || 5 cos(- 37°) = - 3 N

|-

| '''<math>\sum{F}</math>''' || m a = 1 a || 0

|-

|}

Writing the equation for the x-component, 

- 6 + 4 = a 

we obtain <math>a = - 2 m/s^2</math>. 

We don't need to calculate the normal force. 

These findings, together with the equations of kinematics, allow us to compute displacement of the block over the first two seconds. Given the initial velocity <math>v_0 = 4 m/s</math>, we can write the equation for the displacement 

<math>\Delta x = v_{0x} t</math> + ½a<math> t^2</math> = 4 × 2 + ½ × (-2) × 22 = 4 m

Newtons Laws EX 15 - Revision history

Nuzhat at 16:16, 14 April 2011