In Fig.7-49a , a 2.0 N force is applied to a 4.0 kg block at a downward angle $\mathit{\theta }$as the block moves rightward through 1.0 m across a frictionless floor. Find an expression for the speed ${\mathit{v}}_{\mathbf{f}}$of the block at the end of that distance if the block’s initial velocity is (a) 0 and (b) 1.0 m/s to the right. (c) The situation in Fig. 7 - 49b is similar in that the block is initially moving at 1.0 m/s to the right, but now the 2.0 N force is directed downward to the left. Find an expression for the speed ${\mathbf{v}}_{\mathbf{f}}$of the block at the end of the distance. (d) Graph all three expressions for ${\mathbf{v}}_{\mathbf{f}}$versus downward angle for. Interpret the graphs.

It is given that,

The force acting on the block is$F=2\mathrm{N}$.

Mass of the block is$m=4\mathrm{kg}$

Displacement of the object is 1 m .

The problem is based onthe work-energy theorem which states that the net work done by the forces applied on object is equal to the change in its kinetic energy. It alsodeals with the work done. Work, energy are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it.Find the expression for by equating the formulae for work obtained from its definition and from work-energy theorem.

Formulae:

$$\begin{gathered} W=\overrightarrow{F}.\overrightarrow{d} \\ =Fd\cos \theta \\ W=K.E \\ =\frac{1}{2}m{v}^2 \end{gathered}$$

Where,

F is force, dis displacement, K.E.is kinetic energy, v is the velocity, m is the massand Wis the work done.

Now,

$$\begin{gathered} W=\overrightarrow{F}.\overrightarrow{d} \\ =Fd\cos \theta \end{gathered}$$

According to the work-energy theorem,

$$\begin{gathered} W=K.E \\ =\frac{1}{2}m\left(v_f^2-v_i^2\right) \\ \frac{1}{2}m\left(v_f^2-v_i^2\right)=\frac{2}{m}Fd\cos \theta \\ {v}_f^2=v_i^2+\frac{2}{m}Fd\cos \theta \end{gathered}$$

Hence, an expression for the speed$v_f$ of the block if its initial velocity is zero m/s is$v_f=\sqrt{\cos \theta }$.

If initial velocity of the block is 0 m/s,

$$\begin{gathered} v_f^2=0+\frac{2}{4}\left(2\right)\left(1\right)\cos \theta \\ {v}_f=\sqrt{\cos \theta } \end{gathered}$$

If initial velocity of the block is 1 m/s,

$$\begin{gathered} v_f^2=0+\frac{2}{4}\left(2\right)\left(1\right)\cos \theta \\ {v}_f=\sqrt{1+\cos \theta } \end{gathered}$$

Hence, an expression for the speed of the block if its initial velocity is 1 m/s is$v_f=\sqrt{1+\cos \theta }$.

If initial velocity of the block is 1 m/s and force is directed downward to the left,

$$\begin{gathered} v_f=\sqrt{1+\cos \left(180°-\theta \right)} \\ {v}_f=\sqrt{1-\cos \theta } \end{gathered}$$

Hence, an expression for the speed$v_f$ of the block if its initial velocity is 1 m/s and force is directed downward to the left is$v_f=\sqrt{1-\cos \theta }$

Graph for expressions obtained in part a, b, and c is as follow:

Hence, the graph for all three expressions for $v_f$ versus downward angle$\theta$ is plotted.