Q17E - University Physics with Modern Physics

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 15.0N . (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? 

Step-by-Step Solution

Answer

(a) The free body diagrams of brick load and counterweight is given in figure (1) and figure (2).

(b) The acceleration of either block is $3.75 \mathrm{m}/{\mathrm{s}}^2$ .

(c) The mass of the hanging block is 2.48 kg. 

(d) The tension in the rope is less than the weight of hanging block.

Step 1: Given Data:

The mass of block on horizontal surface is $M=4 \mathrm{kg}$ 

The tension in the rope is $T=15 \mathrm{N}$  

Step 2: Acceleration of block:

The acceleration of the block is calculated by considering equilibrium of forces in horizontal direction.

The expression for the force is given by,

$F=ma$  

Here $m$ is the mass,  $F$ is the force and $a$ is the acceleration.

Step 3: Determine the free body diagrams for bricks and counterweight

(a)

Draw the diagram of the system,

                                     

The free body diagram of both the masses is are shown below,

                                                   

Step 4: Determine the acceleration of either block

(b)

The acceleration of the block on horizontal surface is calculated as:

$T=M{a}_x$  

Here, $a_x$ is the acceleration of horizontal block.

Substitute 15N for T  and  4kg for M  in the above equation.

$$\begin{gathered} 15N=\left(4 \mathrm{kg}\right)a_x \\ a_x=3.75 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$ 

Therefore, the acceleration of either block is $3.75 \mathrm{m}/{\mathrm{s}}^2$ .

Step 5: Determine the mass of the hanged block

(c)

The mass of hanged block is calculated as:

$T=m\left(g-a_x\right)$  

Here, g  is the gravitational acceleration and its value is $9.8 \mathrm{m}/{\mathrm{s}}^2, T$  is the tension in the moving rope.

Substitute 15N for $T, 9.8 \mathrm{m}/{\mathrm{s}}^2$ for g and $3.75 \mathrm{m}/{\mathrm{s}}^2$ for $a_x$ in the above equation.

$$\begin{gathered} 15 N=m\left(9.8 \mathrm{m}/{\mathrm{s}}^2 -3.75 \mathrm{m}/{\mathrm{s}}^2\right) \\ m=2.48 \mathrm{kg} \end{gathered}$$ 

Therefore, the mass of hanged block is 2.5 kg .

Step 6: Compare the tension from the weight of the hanging block

(d)

The weight of the hanged block is given as:

$$\begin{gathered} w=mg \\ w=\left(2.48 \mathrm{kg}\right)\left(9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ w=24.3 \mathrm{N} \end{gathered}$$  

The weight of the hanged block $\left(24.3 \mathrm{N}\right)$  is greater than the tension $\left(15 \mathrm{N}\right)$ in the rope.

Therefore, the weight of the hanging block is greater than the tension in the rope.