Q72P In Fig.7-49a , a 2.0 N force is ... [FREE SOLUTION]

Chapter 7: Q72P (page 176)

In Fig.7-49a , a 2.0 N force is applied to a 4.0 kg block at a downward angle $胃$ as the block moves rightward through 1.0 m across a frictionless floor. Find an expression for the speed $v_{f}$ of the block at the end of that distance if the block鈥檚 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 $v_{f}$ of the block at the end of the distance. (d) Graph all three expressions for $v_{f}$ versus downward angle for. Interpret the graphs.

Short Answer

Expert verified

(a)An expression for the speed $v_{f}$ of the block if its initial velocity is zero m/s is $v_{f} = \sqrt{\cos 胃}$

(b)An expression for the speed $v_{f}$ of the block if its initial velocity is 1 m/s is $v_{f} = \sqrt{1 + \cos 胃}$

(c)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 $v_{f} = \sqrt{1 - \cos 胃}$

(d)Graph for all three expressions for $v_{f}$ versus downward angle $胃$ is plotted.

Step by step solution

Given information

It is given that,

The force acting on the block is $F = 2 N$ .

Mass of the block is $m = 4 kg$

Displacement of the object is 1 m .

Determining the concept

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:

$W = \overset{鈫�}{F} . \overset{鈫�}{d} = F d \cos 胃 W = K . E = \frac{1}{2} m v^{2}$

Where,

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

(a) Determining an expression for the speed vf of the block if its initial velocity is zero m/s

Now,

$W = \overset{鈫�}{F} . \overset{鈫�}{d} = F d \cos 胃$

According to the work-energy theorem,

$W = K . E = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) \frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = \frac{2}{m} F d \cos 胃 v_{f}^{2} = v_{i}^{2} + \frac{2}{m} F d \cos 胃$

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

(b) determining an expression for the speed vf of the block if its initial velocity is 1 m/s

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

$v_{f}^{2} = 0 + \frac{2}{4} (2) (1) \cos 胃 v_{f} = \sqrt{\cos 胃}$

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

$v_{f}^{2} = 0 + \frac{2}{4} (2) (1) \cos 胃 v_{f} = \sqrt{1 + \cos 胃}$

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

(c) Determining an expression for the speed vf of the block if its initial velocity is 1 m/s and force is directed downward to the left

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

$v_{f} = \sqrt{1 + \cos (180 掳 - 胃)} v_{f} = \sqrt{1 - \cos 胃}$

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 $v_{f} = \sqrt{1 - \cos 胃}$