please show all work and explainations, finna learn some physics!! 2. Given the

Question

please show all work and explainations, finna learn some physics!!

2. Given the following system on a smooth surface with a pulley of negligible mass: m, 2ms. Determine the acceleration of m. Note: additional space is on the following page. mz Name: Given the following system on a smooth surface with a pulley of negligible mass m,-3m2·8-30 the acceleration of mr 3. Determine m1 mz 4. 5. Given the system in problem # 3, for what value of theta will the system remain static (if any)? Show work. List assumptions in the Pre-lab procedures for example, the string's mass, the stretching of the string

Explanation / Answer

2);from diagram we have m1=2m2.As m2 goes down due to gravity, m1 will move horizontally.

Let us say acceleration of m1 be a1 and that of m2 be a2.

Applying newton's 2nd Law if motion on m2, we have

Net force=mass x acceleration

m2g - T=m2a2. ............ (1)

Apply on m1 we have

T=m1a1.........(2)

as there is no other force in the direction of motion(acceleration ).

Put value of equation 2 in equation 1 we get

m2g - m1a1=m2a2

Put m1=2m2

m2g-2m2a1=m2a2

Cancel out m2

g - 2a1=a2

2a1=g - a2

a1=(g - a2) /2,where g =acceleration due to the gravity.

3):the tension T for m1 will get divided into two components one along x axis, i.e Tcosx and one along y axis i.e Tsinx.

Tsinx will balance weight but Tsin will cause it to move.

Let acceleration of m1 be a1 and that of m2 be a2.

Apply newton's 2nd law on m1 in horizontal we have

Net force =mass x acceleration

Tcosx=m1a1

On m2 we have

m2g - T=m2a2..........(2)

Put value of m1=3m2 we have

Tcosx=3m2a1

From 2 we have T=m2g - m2a2

Put value in above equation we get

(m2g - m2a2) cosx=3m2a1

Cancel out m2

(g-a2)cos(30°)=3a1

(g-a2)cos(30°)/3=a1 i.e acceleration of m1.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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