contestada

A force (F) applied to an object of mass m1 produces an acceleration of 3.00 m/s^2.
The same force applied to a second object of mass m2 produces an acceleration of 1.00 m/s^2.

a) what is the value of the ratio m1/m2?

b) if m1 and m2 are combined into one object, what is its acceleration under the action of the force

Respuesta :

a) using f=ma, F=m1 x 3.00 =3m1
also F=m2 x 1.00 = m2
as the Fs are the same: 3m1=m2
this means that the ratio of m1 to m2 is 3 to 1 as there are 3 m1s for every m2.
b) again using f=ma, the combination of the masses is m1 + m2 or 4m1 as there are 3m1s in m2. so F=4m1 x a = 3m1 (as F is the same for all of them).
a = 3m1/4m1 = 3/4ms^-2
MrRoyal

Force is the product of mass and acceleration

  • The ratio m1/m2 is 1/3
  • The acceleration is 3/4

(a) The ratio m1/m2

Force is calculated as:

[tex]F = ma[/tex]

For the first object, we have:

[tex]F = m_1 \times 3[/tex]

For the second object, we have:

[tex]F = m_2 \times 1[/tex]

The forces on both objects are equal.

So, we have:

[tex]m_1 \times 3 = m_2 \times 1[/tex]

Divide both sides by 3m2

[tex]\frac{m_1}{m_2}= \frac{1}{3}[/tex]

Hence, the ratio m1/m2 is 1/3

(b) The acceleration when the masses are combined

In (a), we have:

[tex]\frac{m_1}{m_2}= \frac{1}{3}[/tex]

Make m2 the subject

[tex]m_2 = 3m_1[/tex]

Recall that:

[tex]F = ma[/tex]

So, we have:

[tex]F =(m_1 +m_2) \times a[/tex]

Substitute [tex]m_2 = 3m_1[/tex]

[tex]F =(m_1 +3m_1) \times a[/tex]

[tex]F =4m_1 \times a[/tex]

Substitute [tex]F = m_1 \times 3[/tex] --- because the force is the same

[tex]m_1 \times 3 =4m_1 \times a[/tex]

Cancel out m1

[tex]3 =4 \times a[/tex]

Divide both sides by 4

[tex]a = \frac 34[/tex]

Hence, the acceleration is 3/4

Read more about force and acceleration at:

https://brainly.com/question/13370981

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