Q6-42E - University Physics with Modern Physics

Question

A $\mathbf{4}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$block of ice is placed against a horizontal spring that has force constant$\mathit{k}\mathbf{=}\mathbf{200}\mathbf{\text{\hspace{0.17em}N/m}}$ and is compressed$\mathbf{0}\mathbf{.025}\mathbf{\text{\hspace{0.17em}m}}$ . The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. 

(a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. 

(b) What is the speed of the block after it leaves the spring?

 

Step-by-Step Solution

Answer

a) The work done on the block by spring from its initial position to where it comes to its original position is$0.0625\text{\hspace{0.17em}J}$ 

b) The velocity of the block after it leaves the spring is $0.18\text{\hspace{0.17em}m/s}$.

 

Step 1: Identification of given data

The given data can be listed below,

• The mass of the block is,$m=4\text{\hspace{0.17em}kg}$.
• The force constant of the spring is,$k=200\text{\hspace{0.17em}N/m}$.
• The compressed length of the spring is,$x_1=0.025\text{\hspace{0.17em}m}$.

Step 2: Concept/Significance of Hooke's law

The well-known equation used to describe a mass on a spring is Hooke's law. It measures tension in a solid within the elastic limit.

Step 3: (a) Determination of the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length

The work done on the blockby spring is given by,

 

$W=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$

 

Here, k is the spring constant,$x_1$ is the compressed distance, and$x_2$ is the final compressed distance whose value is zero as it eventually comes to rest.

 

Substitute all the values in the above,

$\begin{array}{c}W=\frac{1}{2}\left(200\text{\hspace{0.17em}N/m}\right){\left(0.025\text{\hspace{0.17em}m}\right)}^2-\frac{1}{2}\left(200\text{\hspace{0.17em}N/m}\right){\left(0\text{\hspace{0.17em}m}\right)}^2\\ =\left(100\text{\hspace{0.17em}N/m}\right){(-0.025\text{\hspace{0.17em}m})}^2\\ =0.0625\text{\hspace{0.17em}J}\end{array}$

Thus, the work done on the block by spring from its initial position to where it comes to its original position is $0.0625\text{\hspace{0.17em}J}$. 

Step 3: (b) Determination of the speed of the block after it leaves the spring

By the work-energy theorem, the velocity of the block is given by,

 

$\begin{array}{c}W=K_f-K_i\\ =\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2\\ v_2=\sqrt{\frac{2W}{m}}\end{array}$

 

Here, m is the mass of the block, $v_1$is the initial velocity of the block, and $v_2$is the final velocity of the block.

 

Substitute all the values in the above,

 

$\begin{array}{c}W=\sqrt{\frac{2\times 0.0625\text{\hspace{0.17em}J}}{4\text{\hspace{0.17em}kg}}}\\ =0.18\text{\hspace{0.17em}m/s}\end{array}$

 

Thus, the velocity of the block after it leaves the spring is$0.18\text{\hspace{0.17em}m/s}$ .