Q79P - University Physics with Modern Physics

Question

A 5.0 kg block is moving at ${\mathbf{v}}_{\mathbf{0}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{m}\mathbf{/}\mathbf{s}$ along a frictionless, horizontal surface toward a spring with force constant k = 500 N/mk that is attached to a wall (Fig. P6.79). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of ${\mathbf{v}}_{\mathbf{0}}$?

                

Step-by-Step Solution

Answer

1. The maximum distance compressed by the spring is 0.6 m.
2. The velocity of the block after spring compressed to  is 0.150 m is 1.5 m/s .

Step 1: Identification of given data

The given data can be listed below,

• The mass of the block is, m = kg
• The velocity of the block is, ${\mathrm{v}}_0=6.00 \mathrm{m}/\mathrm{s}$
• The force constant of the spring is, k = 500 N/m

Step 2: Concept/Significance of spring constant.

The spring constant, which has a value of N/m and is the ratio of a spring's force to its additional length, is a measurement of the elasticity of the spring force.

Step 3: (a) Determination of the maximum distance the spring will be compressed

According to the work-energy theorem, the maximum compressed distance is given by,

$$\begin{gathered} W\mathit{=}{K}_{\mathit{i}\mathit{n}\mathit{i}\mathit{t}\mathit{i}\mathit{a}\mathit{l}}\mathit{-}{K}_{\mathit{f}\mathit{i}\mathit{n}\mathit{a}\mathit{l}} \\ \frac{1}{2}k{\left(∆l\right)}^{\mathit{2}}\mathit{=}\frac{\mathit{1}}{\mathit{2}}m{v}_{\mathit{0}}^{\mathit{2}} \\ \mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{∆}l\mathit{=}\sqrt{\frac{\mathit{m}}{\mathit{k}}}{v}_{\mathit{0}} \end{gathered}$$

Here, m is the mass of the block, k is the force constant of the spring, and is the velocity of the block.

Substitute all the values in the above,

$$\begin{gathered} ∆\mathrm{l}=\sqrt{\frac{5\mathrm{kg}}{500 \mathrm{N}/\mathrm{m}}}\left(6.0 \mathrm{m}/\mathrm{s}\right) \\ =0.6 \mathrm{m} \end{gathered}$$

Thus, the maximum distance compressed by the spring is 0.6 m.

Step 4: (b) the maximum value of v 0 when the spring is to compress by no more than 0.150 m.

The velocity of the block after a distance compressed by spring is given by,

 $v_0=\sqrt{\frac{k}{m}}∆l$

Here, m is the mass of the block, k is the force constant of the spring, and is the distance compressed of spring whose value is 0.150 m.

Substitute all the values in the above,

$v_0=\sqrt{\frac{500 N/m}{5 kg}}\left(0.150 m\right)$

Thus, the velocity of the block after spring compressed to 0.150 m is 1.5 m/s .