Question

An object that is originally at locationmoves to locationas shown in Figure 6.80. While it is moving it is acted on by a constant force of.

(a) How much work is done on the object by this force? (b) Does the kinetic energy of the object increase or decrease?

(c) A different object moves from locationto location , as shown in Figure 6.81. While it is moving it is acted on by a constant force of . How much work is done on the second object by this force? (d) Does the kinetic energy of the object increase or decrease?

Short answer

(a) -176 J

(b) The kinetic energy of an object decreases.

(c) 176 J

(d) The kinetic energy of an object increases.

Step-by-step solution

Identification of the given data

The given data is listed below as,

• The initial location of the object in the first case is, .
• The final location of the object in the first case is, .
• The force acting on the object is in the first case, .
• The initial location of the object in the second case, .
• The final location of the object in the second case, .

The force acting on the object in the second case is, .

Significance of the work done

The work done on an object is equal to the product of the force exerted by the object and the distance moved by the object.

The expression for the work done is as follows,

W = F d

Here, is the force acting on the object and is the distance moved by the object.

Determination of the work done in the first case

(a)

The equation of the work done on the object by the force in the first case is expressed as,

$$\begin{gathered} W_1=F_1{d}_1 \\ =F_1\left(F_{1f}-d_{1j}\right) \end{gathered}$$

Here,$F_1$ is the force exerted by the object in the first case and $d_{1f}$, and$d_{1j}$ are the final and the initial location of the object in the first case.

For , and .

role="math" localid="1654148746790" $$\begin{gathered} W=\left(\left\{22,0,0\right\}N\right)\times \left(\left\{-25,0,0\right\}m\right)m-\left(\left\{-17,0,0\right\}m\right) \\ =\left(\left\{22,0,0\right\}N\right)\times \left\{-8,0,0\right\}m \\ =176Nm\times \frac{1N}{1Nm} \\ =176N \end{gathered}$$

Thus, the work done by the force on the object in the first case is .

Determination of the kinetic energy of the object in the first case

(b)

The work done of the object is obtained as $-176J$. By work-energy theorem, the work done is equal to the change in the kinetic energy of an object. So, it is found that the value is negative which means that kinetic energy of the object is decrementing.

Thus, the kinetic energy of the object decreases.

Determination of the work done in the second case

(c)

The equation of the work done on the object by the force in the second case is expressed as,

$$\begin{gathered} W_2=F_2{d}_2 \\ =F_2\left(d_{2f}-d_{2i}\right) \end{gathered}$$

Here, F₂is the force exerted by the object in the second case and d_2f , and di are the final and the initial location of the object in the second case.

For

Thus, the work done by the force on the object in the second case is 176 J .

Determination of the kinetic energy of the object in the second case

(d)

The work done of the object is obtained as $176J$. By work-energy theorem, the work done is equal to the change in the kinetic energy of an object. So, it is found that the value is positive which means that kinetic energy of the object is incrementing.

Thus, the kinetic energy of the object increases.