4.3. Newton’s Third Law http://www.ck12.org
3a. Find the mass of the block of ice inFigure4.7, use 9. 8 m/s^2 forg.
Answer:W=mg,m= 94900. 8 m/Ns 2 = 500 kg
3b. Find the acceleration of the block of ice inFigure4.7.
Answer:∑F=Ma,∑F=Fa p=ma.
100 N= ( 500 kg)a,a= 0. 20 m/s^2
3c. Find the velocity of the block att= 100 s
Answer:Vf=at+Vi,( 0. 20 )( 100 )+ 0 = 20 m/s
3d. Find the displacement of the block att= 100 s.
Answer:∆x=^12 (Vi+Vf)t=^12 ( 0 + 20 )( 100 ) = 1000 m
Example 4: A Touching Story
InFigure4.8, BlockAhas a mass of 10.00 kg and BlockBhas a mass of 6.00 kg. Both blocks are in contact with
each other, with BlockAexperiencing an applied 70.0 N force to the right as shown. Note that both blocks have the
same acceleration.
Note: When referring to more than one mass we often use the word “system.”
FIGURE 4.8
4a. Draw the FBD’s for BlockAand BlockB
FIGURE 4.9
Answer:As BlockAmoves to the right it experiences a force from BlockBto the left.