Question

A light rope is attached to a block with mass 4.00 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 15.0 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) pic is attached

b) 3.75 m/s^2

c) m=2.48 kg

d) T=0.617 W

Explanation:

a) pic is attached

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:

∑F_x=m*a_x

a_x=15/4

     =3.75 m/s^2

c) in order to calculate m , we apply newton law on the hanging block

∑F_x=m*a_y

T-W=-m*a_y

T-mg=-m*a_y

T=mg-m*a_y

T=m"(g-a_y)

a_x=a_y

15=m(9.8-3.75)

m=2.48 kg

note:

the sign of a_y is -ve cause a_y is the -ve y-direction and it has the same magnitude of a_x

d)

calculate the weight of the hanging block:

W=mg

W=2.48*9.8

  =24.3 N

T=15/W

T=0.617 W

T<weight of the hanging block