Question

A block with mass mA = 15.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second block with mass mB = 6.0 kg which hangs vertically.

Determine the magnitude of the acceleration of the system.

Express your answer to two significant figures and include the appropriate units.

If initially mA is at rest 1.250 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?

Express your answer to two significant figures and include the appropriate units.

If mB = 1.0 kg, how large must mA be if the acceleration of the system is to be kept at g100?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Let there be a tension T in the wire ...
Since both blocks are connected by cord, so both the blocks will have same acceleration say a.
So for block B,
   (Newton's second law)

w_b - T = 6 * a _______________(1)

Here w_b = weight of block B

= m_b * g   (g = acceleration due to gravity)

= 6 * g
from equation (1)

6* g - T = 6 * a ___________________(2)

Now, applying Newton's second law for block A

  

There is only one force acting on the block A which is tension T .

So,

T = 15 * a _________________(3)

Substituting the value of T in equation (2)

6 * g - 15 * a = 6 * a

6 * g = 21 * a

  

a = 2.80 m/s²

The block A is moving at a constant acceleration of 2.80 m/s² ...Now, using the equation

  

here s = displacement = 1.250 m

u = initial velocity = 0 m/s

a = acceleration = 2.80 m/s²

1.250 = 0 + 0.5 * 2.80 * t²

  

t = 0.94 s

Hence , it would take 0.94 s to reach the edge.

If m_b = 1.0 kg

Therefore , equation (2) become

1 * g - T = 1 * a

Here required acceleration is 0.01g

g - T = 0.01 * g

T = 0.99 * g

acceleration of block B , a = T/m_A

0.01 * g = 0.99 * g /m_A

  

m_A = 99 kg