Q40P - University Physics with Modern Physics

Question

Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block A has mass , and block B has mass . A constant horizontal force F=60.0 N is applied to block A (Fig. P4.40). In the first 5.00 s after the force is applied, block A moves 18.0 m to the right. (a) While the blocks are moving, what is the tension  in the rope that connects the two blocks? (b) What is the mass of block B?

Step-by-Step Solution

Answer

(a) The tension T in the rope that connects the two blocks is 38.4 N.

(b) The mass of block B is 26.6 kg.

Step 1: Identification of the given data

The given data is listed below as:

 

• The mass of the block A is, M=15.0 kg
• The mass of the block B is, m
• The force applied on the block A is, F=60.0 N
• The time taken to apply to force is, t=5.00 s
• The distance moved by the block A is,s=18.0 m 

Step 2: Significance of the force applied

The force on an object is described as the product of the mass and the acceleration of that object. The force and the tension on an object goes is beneficial for understanding the distance moved by the object.

Step 3: (a) Determination of the tension in the rope

The equation of the acceleration of the rope is expressed as:

 

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s-ut=\frac{1}{2}a{t}^2 \\ 2\left(s-ut\right)=a{t}^2 \\ a=\frac{2\left(s-ut\right)}{t^2} \end{gathered}$$

 

Here, s is the distance moved by the block A, u is the initial speed of the blocks, t is the time taken to apply the force and is the acceleration of the rope.

 

As the blocks were at rest initially, then the initial speed is zero.

 

The equation of the tension in the rope is expressed as:

$$\begin{gathered} F-T=Ma \\ T=F-Ma \\ =F-M\frac{2\left(s-ut\right)}{t^2} \end{gathered}$$ 

 

 

Here, F is the force applied on the block A, M is the mass of the block A and T is the tension of the rope.

 

Substitute all the values in the above equation.

 

$$\begin{gathered} \mathrm{T}=60.0 \mathrm{N}-\left(15.0 \mathrm{kg}\right)\frac{2\left(18.0 \mathrm{m}-0\times \mathrm{t}\right)}{{\left(5.00 \mathrm{s}\right)}^2} \\ =60.0 \mathrm{N}-\left(15.0 \mathrm{kg}\right)\frac{\left(36.0 \mathrm{m}\right)}{\left(25.00 {\mathrm{s}}^2\right)} \\ =60.0 \mathrm{N}-\left(15.0 \mathrm{kg}\right)\left(1.44 \mathrm{m}/{\mathrm{s}}^2\right) \\ =60.0 \mathrm{N}-21.6 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

 

Hence, further as:

 

$$\begin{gathered} \mathrm{T}=60.0 \mathrm{N}-21.6 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =60.0 \mathrm{N}-21.6 \mathrm{N} \\ =38.4 \mathrm{N} \end{gathered}$$ 

 

Thus, the tension T in the rope that connects the two blocks is 38.4 N.

Step 4: (b) Determination of the mass of the block B

The equation of the mass of the block B is expressed as:

 

F-T=-ma

 

Here, m is the mass of the block B, a is the acceleration that is directed upwards and T is the tension of the rope.

 

As there are no force acts on the block B, hence the force on the block B is zero.

 

Hence, the above equation can be expressed as:

 

$$\begin{gathered} 0-T=-ma \\ T=ma \\ m=\frac{T}{a} \end{gathered}$$

 

Substitute the value of the equation (i) in the above equation.

 

$m=\frac{T}{{\displaystyle \frac{2\left(s-ut\right)}{t^2}}}$ 

 

Substitute all the values in the above equation.

 

$$\begin{gathered} \mathrm{m}=\frac{38.4 \mathrm{N}}{2\left(18.0 \mathrm{m}-\left(0\right)\mathrm{t}\right)}{\left(5.00 \mathrm{s}\right)}^2 \\ =\frac{38.4 \mathrm{N}}{\left(36.0 \mathrm{m}\right)}\left(25.00 {\mathrm{s}}^2\right) \\ =\left(1.06 \mathrm{N}/\mathrm{m}\right)\left(25.00 {\mathrm{s}}^2\right) \\ =26.6 \mathrm{N}.{\mathrm{s}}^2/\mathrm{m} \end{gathered}$$ 

 

Hence, further as:

 

$$\begin{gathered} \mathrm{m}=26.6 \mathrm{N}.{\mathrm{s}}^2/\mathrm{m}\times \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}} \\ =26.6 \mathrm{kg}.\mathrm{m}.{\mathrm{s}}^2/\mathrm{m}.{\mathrm{s}}^2 \\ =26.6 \mathrm{kg} \end{gathered}$$

 

Thus, the mass of block B is 26.6 kg.