Question

Figure shows a 5.00 kgblock being pulled along a frictionless floor by a cord that applies a force of constant magnitude20.0Nbut with an anglethat varies with time. When angle $\mathit{\theta }\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}$, (a) at what rate is the acceleration of the block changing if$\mathit{\theta }\left(t\right)\mathbf{=}\left(2.00\times {10}^{-2}deg/s\right){\mathit{t}}^{\mathbf{-}\mathbf{2}}$and (b) at what rate is the acceleration of the block changing if $\mathit{\theta }\left(t\right)\mathbf{=}\left(2.00\times {10}^{-2}deg/s\right)\mathit{t}$?

(Hint:The angle should be in radians.)

Short answer

a) Rate of change of acceleration when $\theta \left(t\right)=\left(2.00\times {10}^{-2}\frac{deg}{s}\right)$is

$\frac{d{a}_{\times }}{dt}=-5.90\times {10}^{-4}\frac{m}{s^3}$

b) Rate of change of acceleration when$\theta \left(t\right)=\left(2.00\times {10}^{-2}\frac{deg}{s}\right)$

role="math" localid="1660901068388" $\frac{d{a}_{\times }}{dt}=5.90\times {10}^{-4}\frac{m}{s^3}$

Step-by-step solution

Given information

1. Mass of a block,$m=5.00kg$
2. Force,$F=20.0N$
3. $\theta =25.0°$
   

Understanding the concept of Newton’s law

The net force is equal to the product of mass and acceleration. An acceleration is time rate of change of velocity.

Use the net force concept for the rate of change of acceleration. By using a derivative, get the equation for the rate of change of acceleration.

Formula:

$$\begin{gathered} \mathit{F}\mathbf{=}\mathit{m}\mathit{a} \\ \frac{\mathbf{d}\mathbf{a}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{m}}\frac{\mathbf{d}\mathbf{f}}{\mathbf{d}\mathbf{t}} \end{gathered}$$

Here, F is net force, m is mass of the object,a is acceleration.

(a) Calculate the rate of change of acceleration

By applying Newton’s second law to the horizontal axis, we get

$$\begin{gathered} F\cos \theta =m{a}_x \\ {a}_x=\frac{F\cos \theta }{m} \\ \frac{d{a}_x}{dt}=\frac{F\sin \theta }{m}\frac{d\theta }{dt} \end{gathered}$$

We have,

$$\begin{gathered} \theta \left(t\right)=\left(2.00\times {10}^{-2}\frac{deg}{s}\right)t \\ \frac{d\theta }{dt}=\left(2.00\times {10}^{-2}\frac{deg}{s}\right)\left(\frac{\mathrm{\pi rad}}{180°}\right) \\ \frac{d\theta }{dt}=3.49\times {10}^{-4}\frac{rad}{s} \end{gathered}$$

Use the given value in the above equation to get the rate of change of acceleration.

$$\begin{gathered} \frac{d{a}_x}{dt}=-\frac{20.0\times \sin 25.0}{5.00}\left(3.49\times {10}^{-4}\frac{rad}{s}\right) \\ \frac{d{a}_x}{dt}=-5.90\times {10}^{-4}\frac{m}{s^3} \end{gathered}$$

Therefore, the rate of change of acceleration is $5.90\times {10}^{-4}\frac{m}{s^3}$.

(b) Calculate the rate of change of acceleration

We have,

$$\begin{gathered} \theta \left(t\right)=-\left(2.00\times {10}^{-2}\frac{deg}{s}\right)t \\ \frac{d\theta }{dt}=-\left(2.00\times {10}^{-2}\frac{deg}{s}\right)\left(\frac{\mathrm{\pi rad}}{180°}\right) \\ \frac{d\theta }{dt}=-3.49\times {10}^{-4}\frac{rad}{s} \end{gathered}$$

We can use the given value in the above equation to get the rate of change of acceleration.

localid="1660902015077" $$\begin{gathered} \frac{d{a}_x}{dt}=-\frac{20.0\times \sin 25.0}{5.00}\left(-3.49\times {10}^{-4}\frac{rad}{s}\right) \\ \frac{d{a}_x}{dt}=5.90\times {10}^{-4}\frac{m}{s^3} \end{gathered}$$

Therefore, the rate of change of acceleration is $5.90\times {10}^{-4}\frac{m}{s^3}$.