Acceleration of the lower block

In summary: You should get a final answer of 2.65 m/s^2.In summary, the lower block in the diagram experiences a force of 58.6 N and has a mass of 4.68 kg. The upper block has a mass of 2.11 kg and both blocks have a coefficient of kinetic friction of 0.204. Using the equations of Newton's second law and the coefficient of friction, the acceleration of the lower block is found to be 2.65 m/s^2.
  • #1
brunettegurl
138
0

Homework Statement



In the diagram shown below, the lower block is acted on by a force, F, which has a magnitude of 58.6 N. The coefficient of kinetic friction between the lower block and the surface is 0.204. The coefficient of kinetic friction between the lower block and the upper block is also 0.204. What is the acceleration of the lower block, if the mass of the lower block is 4.68 kg and the mass of the upper block is 2.11 kg?

Homework Equations



F=ma / friction = [tex]\mu[/tex]mg

The Attempt at a Solution



so i wrote the forces that affect each box

For the upper box ::[tex]\sum[/tex] F= T-[tex]\mu[/tex]m1g
for the lower box :: [tex]\sum[/tex] F= F-T- friction1- friction2

i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what I'm doing wrong thanks :))
 

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  • #2
brunettegurl said:
so i wrote the forces that affect each box

For the upper box ::[tex]\sum[/tex] F= T-[tex]\mu[/tex]m1g
for the lower box :: [tex]\sum[/tex] F= F-T- friction1- friction2
OK, but what did you use for the friction forces on the lower box?
i solved the first equation for tension and then substituted it into the 2nd equation and i solved for acceleration...can u pls. tell me what I'm doing wrong thanks :))
I assume you applied Newton's 2nd law to each box to get two equations. Show exactly what you did and then we can see what went wrong.
 
  • #3
for friction 1 i used (coeffecient of frictionxmass1xgravity) and for friction 2 i use (coeffecient of frictionxmass2xgravity)
and the equation i got before isolating for acceleration is ::

a(m2+m1)= Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity)

so then a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)
 
  • #4
brunettegurl said:
for friction 1 i used (coeffecient of frictionxmass1xgravity)
OK.
and for friction 2 i use (coeffecient of frictionxmass2xgravity)
Careful: What's the normal force between the bottom block and the floor?
 
  • #5
is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...?
 
  • #6
brunettegurl said:
is it f 2 = coeffecient of frictionxmass2+ mass1xgravity...?
Yes, the normal force is the combined weight of both blocks, thus f2 = μ(m1 + m2)g.
 
  • #7
ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??
 
  • #8
brunettegurl said:
ok so i changed that and my answer is still coming out wrong is there anything else wrong i cld have done in the initial equation??
Show me your final equation.
 
  • #9
my final equation wld look like ::
m2a = [Fapplied-m1a-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)]


a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)

Fapplied is the force given in the question
 
  • #10
brunettegurl said:
my final equation wld look like ::
m2a = [Fapplied-m1a-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)]


a = [Fapplied-(coeffecient of frictionxmass1xgravity)-(coeffecient of frictionxmass1=mass2xgravity)-(coeffecient of frictionxmass2xgravity)] /(m1+m2)

Fapplied is the force given in the question
Your equations are a bit confusing, due to what may be a few typos. (The '=', for one.) Can you rewrite using F, m1, m2, μ, and g? Try to simplify as much as possible.
 
  • #11
m2a= [F -m1a- (μm1g)- (μg(m1+m2)) - (μm2g)]

a = [F - (μm1g) - (μg(m1+m2)) - (μm2g)]/(m1+m2)

iss that clearer ??
 
  • #12
brunettegurl said:
m2a= [F -m1a- (μm1g)- (μg(m1+m2)) - (μm2g)]

a = [F - (μm1g) - (μg(m1+m2)) - (μm2g)]/(m1+m2)

iss that clearer ??
Yes, much clearer. You have an extra m2 term in there for some reason. If you write the original Newton's law equations for m1 and m2, perhaps we can see where it got snuck in.
 
  • #13
this is what i have as my equations for m1 and m2

m1:: m1a= T- μm1g
m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g

i don't see the extra m2 in my equations can u be a bit more specific
 
  • #14
brunettegurl said:
this is what i have as my equations for m1 and m2

m1:: m1a= T- μm1g
m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g
That second friction term in the m2 equation is the problem--you're using the wrong mass.
 
  • #15
so this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm2g wld be this m2 :: m2a = Fapp - T- μ(m1+m2)g- μm1g...
then the final equation wld look like a = [Fapp - (μm1g) - (μg(m1+m2)) - (μm1g)]/(m1+m2)
is that correct??
 
  • #16
brunettegurl said:
then the final equation wld look like a = [Fapp - (μm1g) - (μg(m1+m2)) - (μm1g)]/(m1+m2)
is that correct??
That looks good. But consolidate some of those terms.
 

1. What is "Acceleration of the lower block"?

"Acceleration of the lower block" refers to the rate of change of velocity of a lower block in a system of objects. It is a measure of how quickly the lower block's speed is changing.

2. How is acceleration of the lower block calculated?

Acceleration of the lower block can be calculated by dividing the change in velocity by the time it takes for the change to occur. This can be represented by the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What factors affect the acceleration of the lower block?

The acceleration of the lower block can be affected by several factors, such as the mass of the block, the force acting on the block, and the friction between the block and its surface. The direction of the force and the angle of the surface can also impact the acceleration.

4. How does Newton's Second Law relate to the acceleration of the lower block?

According to Newton's Second Law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This means that a larger force or a smaller mass will result in a greater acceleration of the lower block.

5. Can the acceleration of the lower block ever be negative?

Yes, the acceleration of the lower block can be negative if the force acting on it is in the opposite direction of its initial velocity. This means that the block is slowing down instead of speeding up.

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