Question

Blocks A and B areconnected by a massless stringof fixed length. The string slides over a pulley without any friction. Block A has a mass of 2??and block B has a mass of 1??, as shown. The blocks have just this instant been released from rest.

a.Will the speed of block A be greater than, equal to, or less than the speed of block B? Will the blocks accelerate? If yes, will the blocks have the same magnitude of acceleration, or will one be greater than the other?

b.Draw a separate free body diagram for each block. Be sure vector lengths indicate the relative sizes of the forces.(Note: the stringpullsupwardon each block with the same magnitude of force.)

c.Rank in order, from largest to smallest,allthe vertical forces. Explain your ranking.

d.Use Newton’s second law with your free body diagrams to determine the acceleration of the blocks and the force exerted by the rope.

e.How long does block A take to reach the floor from a height of 1.5?? How fast is block A going when it strikes the floor? (We assumethe rope is sufficiently long.)Drawing a graph of velocity vs. time may be helpful.

Answer 1

see the diagram :

2kg ing

a) As the length of the string does not change, the speed of the blocks and magnitude of acceleration of the blocks is same.

b) see the diagram for A and B separately :

AT 2kg A va G ing 24 ig

T = tension in string

2g = weight of 2 kg block

1g = weight of 1 kg block

c) As the pulley does not rotate, the tension in the string on both the sides is same.

weight of 2 kg block is greater than weight of 1 kg block.

d) net force on block A is downward :

>> 2* 9.81 - T = 2a

$\Rightarrow 19.62 - T = 2a$-----(i)

net force on block B is upward :

$T-1g=1a$

$\Rightarrow T-9.81=a$ -----(ii)

solving the equations of (i) and (ii) :

$a = \frac{19.62-9.81}{3} = 3.27 m/s^{2}$ [answer]

e) distance of fall = h = 1.5 m

using kinematic equation :

$s = ut + \frac{1}{2}at^{2}$

$\Rightarrow 1.5 = 0 + \frac{1}{2}*3.27*t^{2}$

$\Rightarrow t = \sqrt{\frac{2*1.5}{3.27}}=0.96 s$ [answer]

speed of block A be v.

using kinematic equation :

v = u + at = 0 + 3.27 * 0.96 = 3.14 m/s. [answer]