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

Answer:

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

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

Answer 2

Answer:

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

b) W=(3/2)mgd

c) a=g/2

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:

$-mg+3mg=ma+3ma\\\\2mg=4ma\\\\a=\frac{g}{2}$

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

$v^2=v_0^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{2ad}=\sqrt{gd}$

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

$W_n=-Td+(3m)gd\\\\T=ma+mg=m(\frac{g}{2}+g)=\frac{3}{2}mg\\\\W_n=-\frac{3}{2}mgd+3mgd=\frac{3}{2}mgd$

c) a=g/2