The Atwood Machfne Free Body Diagrams for Mi and M2 Puttey M1 M2g M19 Ma Equatio

Question

The Atwood Machfne Free Body Diagrams for Mi and M2 Puttey M1 M2g M19 Ma Equations of Motion: M2 7.2.1 Which will have greater value of acceleration? Select One 7.2.2 Why? 7.2.3 What will be the direction of the velocity in M1 (blue)? Select One 7.2.4 What will be the direction of the acceleration in M1 (blue)? Select One 7.2.5 what will be the direction of the velocity in M2 (orange)? ( Select One 7.2.6 What will be the direction of the acceleration in M2 (orange)? Select One 7.2.7 Compute the value of the acceleration in either mass. Suppose M1 is 200 grams and M2 is 70 gr. As you can see in the picture, there are 2 equations for 2 unknowns. Final hint: Both masses have the same acceleration. Show your work.

Explanation / Answer

7.2.1)   The acceleration is same for both masses

7.2.2) The tension in the string does not change during its motion so the acceleration remains same

7.2.3) The direction of the velocity in M1 is Downward

Because the net force on M1 is Downward that is the weight of M1 is greater then tension.

7.2.4) The direction of the acceleration in M1 is Downward

Because the velocity increases in downward direction

7.2.5) The direction of the velocity in M2 is Upward

Because the net force on M1 is Upward that is the tension is greater than weight.

7.2.6) The direction of the acceleration in M2 is Upward

Because the velocity increases in Upward direction.

7.2.7)

We have the equations

(M1 * g) - T = M1 *a and

T - (M2 * g) = M2 * a

Adding these two equations we get

(M1 - M2) * g = (M1 + M2) * a ------(1)

Given M1 = 200 grams = 0.2 kg

M2 = 70 grams = 0.07 kg

we get from eq(1) as

(0.2 - 0.07) * 9.8 = (0.2 + 0.07) * a

a = 4.72 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.