Two blocks are connected by a very light string passing over a massless and frictionless pulley (Fig. E6.7). Traveling at constant speed, the 20.0-N block moves 75.0 cm to the right and the 12.0-N block moves 75.0 cm downward. How much work is done (a) on the 12.0-N block by (i) gravity and (ii) the tension in the string? (b) How much work is done on the 20.0-N block by (i) gravity, (ii) the tension in the string, (iii) friction, and (iv) the normal force? (c) Find the total work done on each block.

The weight of the first body is $W_1=20\mathrm{N}$.

The weight of the second body is $W_2=12\mathrm{N}$.

The distance the bodies travel at a constant speed is $s=0.75\mathrm{m}$.

The work done can be defined as the scalar product of force and the displacement. It is given by,

$$\begin{gathered} \mathit{W}\mathbf{=}\overrightarrow{\mathbf{F}}\mathbf{·}\overrightarrow{\mathbf{s}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{F}\mathbf{·}\mathit{s}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha } \end{gathered}$$

Here,$\overrightarrow{\mathbf{F}}$ is force,$\overrightarrow{\mathbf{s}}$ is displacement, and$\mathit{\alpha }$ is the angle between$\overrightarrow{\mathbf{F}}$ and $\overrightarrow{\mathbf{s}}$.

If the vectors are parallel, then the angle will be $\mathbf{0}\mathbf{°}$, and the work done will be equal to:

$$\begin{gathered} \mathit{W}\mathbf{=}\mathit{F}\mathbf{·}\mathit{s}\mathbf{cos}\left(0°\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{F}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

If the vectors are anti-parallel, then the angle will be $\mathbf{180}\mathbf{°}$, and the work done will be equal to:

$$\begin{gathered} \mathit{W}\mathbf{=}\mathit{F}\mathbf{·}\mathit{s}\mathbf{cos}\mathbf{(}\mathbf{180}\mathbf{°}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathit{F}\mathit{s}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

(a)

If the bodies move at a constant speed, their acceleration is equal to zero hence, the resultant force along the direction of motion will be equal to zero. If observing the body that moves vertically downward, the tension force will be equal to the weight of that body.

$$\begin{gathered} T=W_2 \\ =12\mathrm{N} \end{gathered}$$

The same can be applied to the body moving on a horizontal surface. The weight of the body is acting along the vertical axis, as well as the normal reaction force. Since this body moves at a constant speed, equate the weight of the body with the normal reaction force.

$$\begin{gathered} n=W_1 \\ =20\mathrm{N} \end{gathered}$$

As the body kept on the table moves with constant velocity, the acceleration must be zero. Thus, all the forces acting on the body should be balanced. Therefore the tension force should be equal to the frictional force, for both the forces to be balanced.

$$\begin{gathered} T=f_k \\ =12\mathrm{N} \end{gathered}$$

(i)

The work done by the gravitational force upon the second body can be determined by using equation (1).

$$\begin{gathered} W_{w_2}=F·s \\ =W_2·s \\ =\left(12\mathrm{N}\right)\times \left(0.75\mathrm{m}\right) \\ =9\mathrm{J} \end{gathered}$$

Therefore, the work done by the gravitational force upon the second body is 9 J.

(ii)

The work done by the tension in the string upon the second body can be determined by using equation (2).

$$\begin{gathered} W_{w_2}=F·s \\ =-T·s \\ =\left(-12\mathrm{N}\right)\left(0.75\mathrm{m}\right) \\ =-9 \end{gathered}$$

Therefore, the work done by the tension in the string upon the second body is $-9\mathrm{J}$.

(b)

(i)

As gravity acts in the direction perpendicular to that of motion of the body $\left(\theta =90°\right)$

$$\begin{gathered} W_{W_1}=F·s·\cos \theta \\ =W_1·s·\cos 90° \\ =\left(12\mathrm{N}\right)\left(0.75\mathrm{m}\right)\left(0\right) \\ =0\mathrm{J} \end{gathered}$$

(ii)

Find the work done by the tension in the string.

$$\begin{gathered} W_T=F·s \\ =\left(12\mathrm{N}\right)\left(0.75\mathrm{m}\right) \\ =9\mathrm{J} \end{gathered}$$

(iii)

As mentioned earlier, the frictional force must be equal and opposite to the tension in the string.

$$\begin{gathered} W_f=F·s \\ =\left(-12\mathrm{N}\right)\left(0.75\mathrm{m}\right) \\ =-9\mathrm{J} \end{gathered}$$

(iv)

Find the work done by the normal force.

$$\begin{gathered} W_n=F·s·\cos \theta \\ =\left(20\mathrm{N}\right)\left(0.75\mathrm{m}\right)\cos \left(90°\right) \\ =0\mathrm{J} \end{gathered}$$

Therefore, the required work done are 0 J , 9 J , $-9\mathrm{J}$and 0 J .

(c)

The total work done on 20 N block is-

$$\begin{gathered} {\mathrm{W}}_{\mathrm{total}}=+9\mathrm{J}+\left(-9\mathrm{J}\right)+0+0 \\ =0\mathrm{J} \end{gathered}$$

The total work done on 12 N block is-

$$\begin{gathered} {\mathrm{W}}_{\mathrm{total}}=+9\mathrm{J}+\left(-9\mathrm{J}\right) \\ =0\mathrm{J} \end{gathered}$$

Therefore, the total work done on each block is 0 J .