Question

A light rope is attached to a block with mass 4.10 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 14.7 N.
a. Draw two free-body diagrams: one for each block.
b. What is the acceleration of either block?
c. Find m.
d. How does the tension compare to the weight of the hanging block?

Answer 1

Answer:

a)  please find the attachment

(b) 3.65 m/s^2

c) 2.5 kg

d) 0.617 W

T<weight of the hanging block

Explanation:

a) please find the attachment

(b) Let +x be to the right and +y be upward.

The magnitude of acceleration is the same for the two blocks.  

In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:  

∑Fx=ma_x

   T=m1a_x

  14.7=4.10a_x

 a_x= 3.65 m/s^2

c) in order to calculate m we will apply newton second law on the hanging  

   block

∑F=ma_y

T-W= -ma_y

T-mg= -ma_y

T=mg-ma_y

T=m(g-a_y)

a_x=a_y

14.7=m(9.8-3.65)

 m = 2.5 kg

the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax

d) calculate the weight of the hanging block :

W=mg

W=2.5*9.8

  =25 N

T=14.7/25

 =0.617 W

T<weight of the hanging block