A small block of mass Mb=5.0 kg is placed on a long slab of mass Ms=3.0 kg as sh
ID: 1700918 • Letter: A
Question
A small block of mass Mb=5.0 kg is placed on a long slab of mass Ms=3.0 kg as shown above. Initially, the slab is at rest and the block has a speed of v0=4.0 m/s to the right. The coefficient of kinetic friction between the block and the slab is 0.20, and there is no friction between the slab and the horizontal surface on which it moves.
a) On the dots below that represent the block and slab draw labeled vectors to indicate the forces acting on the slab.
At some moment later, before the block reaches the right end of the slab, both the slab and the block attain identical apeeds, vf.
b) Calculate vf.
c) Calculate the distance the slab has traveled at the moment it reaches vf.
d) Calculate the work done by friction on the slab from the beginning of its motion until it reaches vf.
Explanation / Answer
For the slab
By Newton's 3rd Law: Fn = -Fn' and Ff = - Ff' In otherwords, the reaction force of the friction of the block on the slabcauses the slab to accelerate forward.
How to understand this more intuitively? Suppose you arestanding on a very large sled placed on ice. You exert a backward force on the sled, friction exerts aforward force on your feet. But the reaction forceof *friction* on the sled causes it to slip backward on theice. That ends up being the same direction as you move yourfeet.
(b) First calculate the acceleration of theblock.
ablock = Fblock /m block
= - umblockg / mblock
= - ug
=- (0.20 * 9.8) = - 1.96 m/s2
Then by Newton's 3rd Law, as above, the slab accelerates asfollows:
aslab = Fslab /m slab
= +umblockg / m slab
= + (0.20 *9.8) * 0.50 / 3.0
= 0.327m/s2
Now we have two equations in 2 unknowns
vf2 = vi2 +2abd (block)
vf2 = 0 + 2asd (slab)
Then: v f2 = vi2 +2ab*(vf2 / 2as)
v f2= 4.02 + ab /as vf2
v f2 = 16 + (- 1.97 / 0.327) vf2
v f2 = 16 - 6 vf2
7v f2 = 16
vf = (16/7) = 1.5m/s
(c) We can get distance from one theslab equation:
d =vf2 / 2as = 1.512 /2*0.327
= 3.5 m
As a check-step: Let's put this into the blockequation
vi = vf2 - 2abd = 1.52 +2*1.96*3.5 = 16 m2 / s2 sovi = 4.0 m/s. Good we're on the right track.
(d) Work done will be negative since the force offriction opposes the displacement:
Wf = Ff * d
= - umg * d
= - 0.20 * 0.50 kg * 9.8 * 3.5m
= - 3.4J