Q78P - University Physics with Modern Physics

Question

One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 2.00 - kg block of frictionless ice is attached to the other end and rests on the floor. The spring is initially neither stretched nor compressed. A constant horizontal force of 54.0 N is then applied to the block, in the direction away from the post. (a) What is the speed of the block when the spring is stretched 0.400 m ? (b) At that instant, what are the magnitude and direction of the acceleration of the block?

Step-by-Step Solution

Answer

1. The speed of the block is 3.94 m/s . 
2. The direction of the acceleration is away from the post and the magnitude of the acceleration is 11.8$\mathrm{m}/{\mathrm{s}}^2$.

Horizontal spring executes simple harmonic motion. When the mass M passes through its mean position, an object of mass m is put on it and the two move together.

Step 1: Given data

Force constant = 76.0 N  

Block =  2.00 kg

Constant horizontal force = 54.0 N 

Step 2: (a)Find the speed of the block

The work-energy theorem when applied on the block is,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{tot}}={\mathrm{K}}_2-{\mathrm{K}}_1 \\ {\mathrm{W}}_{\mathrm{F}}+{\mathrm{W}}_{\mathrm{spring}}=\frac{1}{2}\mathrm{mv}{\text{'}}^2-\frac{1}{2}{\mathrm{mv}}^2 \\ {\mathrm{F}}_{\mathrm{x}}+\left(-\frac{1}{2}{\mathrm{kx}}^2\right)=\frac{1}{2}\mathrm{mv}{\text{'}}^2-\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2$ for g , 2.00 kg  for m , 76.0 N  for k , 54.0 N  for F , 0.400 m  for x , and 0 m/s for  v.

 

$$\begin{gathered} \left\{\left(54.0\mathrm{N}\right)\left(0.400 \mathrm{m}\right)-\frac{1}{2}\left(76.0 \mathrm{N}/\mathrm{m}\right){\left(0.400 \mathrm{m}\right)}^2\right\} \\ \mathrm{v}\text{'}=\frac{\sqrt{\left\{\left(54.0 \mathrm{N}\right)\left(0.400 \mathrm{m}\right)-{\displaystyle \frac{1}{2}}\left(76.0 \mathrm{N}/\mathrm{m}\right){\left(0.400 \mathrm{m}\right)}^2\right\}}}{{\displaystyle \frac{1}{2}}\left(2.00 \mathrm{kg}\right)} \\ \mathrm{v}\text{'}=3.94 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Hence, the speed of the block is  3.94 m/s 

Step 3: (b) Find the magnitude and direction of the acceleration of the block

With the assumption that a positive direction is away from the post, Newton’s second law gives,

 

$$\begin{gathered} \mathrm{F}-{\mathrm{F}}_{\mathrm{spring}}=\mathrm{ma} \\ \mathrm{F}-\mathrm{kx}=\mathrm{ma} \end{gathered}$$ 

 

Here, a is the acceleration of the block.

 

Substitute $9.8\mathrm{m}/{\mathrm{s}}^2 \mathrm{for} \mathrm{g}, 2.00\mathrm{kg} \mathrm{for} \mathrm{m}, 76.0 \mathrm{N} \mathrm{for} \mathrm{k} , 54.0 \mathrm{N} \mathrm{for} \mathrm{F} , 0.400 \mathrm{m} \mathrm{for} \mathrm{x}$

$$\begin{gathered} \mathrm{a}=\frac{54.0 \mathrm{N}-\left(76.09\mathrm{N}/\mathrm{m}\right)\times \left(0.400 \mathrm{m}\right)=\left(2.00 \mathrm{kg}\right)\mathrm{a}}{\left(2.00 \mathrm{kg}\right)} \\ \mathrm{a}=11.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The positive sign indicates that the direction of acceleration is away from the post.

 

Hence, the direction of the acceleration is away from the post, and the magnitude of the acceleration is  $11.8\mathrm{m}/{\mathrm{s}}^2$.