Question

Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown above. The pulley has negligible mass and friction and is attached to the ceiling by a bracket. The blocks are simultaneously released from rest.

(a) Derive an equation for the speed v of the block of mass 3m after it falls a distance d in terms of m, d, and physical constants, as appropriate.
(b) Determine the work done by the string on the two-block system as each block moves a distance d .
(c) The acceleration of the center of mass of the blocks-string-pulley system has magnitude aCOM . Briefly explain, in terms of any external forces acting on the system, why aCOM is less than g .

Answer 1

a) v= (g*d)^(1/2)

b) W=(3/2)mgd

c) a=g/2

Step-by-step explanation:

(a) you first use the second Newton law to determine the dynamic equation for each block (you assume that the acceleration is the same for both blocks):

`T_1-W_1=m_1 a=ma\\\\T_2-W_2=-m_2a=-(3m)a`

You can assume T1=T2=T because the pulley has a negligible friction and mass. Then from the last two equation you can obtain a:

`T-mg=ma\\\\T-3mg=-3ma`

you take the difference between the last two equations:

`-mg+3mg=ma+3ma\\\\2mg=4ma\\\\a=(g)/(2)`

for the equation of the velocity of the second block you use:

`v^2=v_0^2+2ad\\\\v_o=0m/s\\\\v=√(2ad)=√(gd)`

b) The work done is given by the following expression:

`W_n=-Td+(3m)gd\\\\T=ma+mg=m((g)/(2)+g)=(3)/(2)mg\\\\W_n=-(3)/(2)mgd+3mgd=(3)/(2)mgd`

c) a=g/2

Answer 2

a) v = √ g x , b) W = 2 m g d , c) a = ½ g

Step-by-step explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

T-W₁ = m a

W₃ - T = M a

w₃ - w₁ = (m + M) a

a = (3m - m) / (m + 3m) g

a = 2/4 g

a = ½ g

the speed of the blocks is

v² = v₀² + 2 ½ g x

v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

W₁ = -W d = - mg d

3m heavy block

W₂ = W d = 3m g d

the total work is

W = W₁ + W₂

W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

a = ½ g