Blocks on a Pulley: Friction & Acceleration

[jti3066]

Homework Statement

A 5.0 kg block on a table is attached to a 4.0 kg block by a light string which passes over a light frictionless pulley. The coefficient of kinetic friction between block 1 and the table is U_k= 0.30. At time t = 0, block 1 is moving toward the pulley with a speed of 0.1 m/s.

A) What is the frictional force acting on Block 1?

B)What is the acceleration of Block 1?

Homework Equations

F_net = ma

F_f = u_k(m_1)(g)

The Attempt at a Solution

A) F_f = 0.30 * 5 * 9.81 = 14.72 N

B) T_1 = T_2

T_1 - u(m_1)g = m_1(a)

T_2 - u(m_1)g = m_1(a)

m_2(g) - u(m_1)g = m_1(a)

a = (((m_2)g)/m_1) - ug = 4.905 m/s^2

 

[tiny-tim]

hi jti3066!

(have a mu: µ and try using the X² and X₂ icons just above the Reply box )

A) F_f = 0.30 * 5 * 9.81 = 14.72 N

B) T_1 = T_2

T_1 - u(m_1)g = m_1(a)

T_2 - u(m_1)g = m_1(a)

m_2(g) - u(m_1)g = m_1(a)

a = (((m_2)g)/m_1) - ug = 4.905 m/s^2

A) is right

but i don't understand your B)

 

[jti3066]

Thanks...

Well, for part be I set up the F_net equation for block one, and the tension from the block to the pully is equal that of the tension in block 2...ie tension in the string...So the first equation I said, "The tension in T_1 minus the force of friction equals the mass 1 times acceleration". From their I know that T_1 = T_2, and T_2 equals mass 2 times g.

Substitution and then sovle for a.

 

[tiny-tim]

hi jti3066!

… T_2 equals mass 2 times g.

oh i see!

no, how can T₂ = m₂g?

m₂ is accelerating!

 

[jti3066]

duh...T_2 = m_2(a)

 

[jti3066]

and then...

a = u(m_1)g/(m_2 - m_1)

 

[tiny-tim]

duh...T_2 = m_2(a)

no! …

T₂ - m₂g = m₂a

(F_total = ma)

 

[jti3066]

B) a = [m_2(g) - um_1(g)]/(m_1 - m_2)

 

[tiny-tim]

B) a = [m_2(g) - um_1(g)]/(m_1 - m_2)

almost right

useful tip …

you only have to look at that to see it must be wrong …

if m₁ = m₂, that would make the acceleration infinite!

 

[jti3066]

Ok i give up...what would be the proper equation and why...Please

 

[tiny-tim]

hi jti3066!

(just got up :zzz: …)

you've made a mistake in a ± sign somewhere …

go back through your equations and see where it was!

 

[jti3066]

B) T_1 - um_1g = m_1a

m_2g - T_2 = m_2a

a = [(m_2g - um_1g)]/(m_1+m_2)

 

[tiny-tim]

Yup!

(btw, you could have got the same result by treating everything as being in a line, and using F = ma on the two blocks as a single body, with force m₂g to the left, and µm₁g to the right! )