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Physics / Physics with MasteringPhysics 4 / Chapter 8 / Problem 60P

Physics with MasteringPhysics | 4th Edition | ISBN: 9780321541635 | Authors: James S. Walker

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A block of mass m = 0.95 kg is connected to a spring of

Chapter 8, Problem 60P

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QUESTION:

Problem 60P

A block of mass m = 0.95 kg is connected to a spring of force constant k = 775 N/m on a smooth, horizontal surface. (a) Plot the potential energy of the spring from x = −5.00 cm to x = 5.00 cm. (b) Determine the turning points of the block if its speed at x = 0 is 1.3 m/s.

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QUESTION:

Problem 60P

A block of mass m = 0.95 kg is connected to a spring of force constant k = 775 N/m on a smooth, horizontal surface. (a) Plot the potential energy of the spring from x = −5.00 cm to x = 5.00 cm. (b) Determine the turning points of the block if its speed at x = 0 is 1.3 m/s.

ANSWER:

Step 1 of 2

Part a

We are required to plot the potential energy of the spring for the given range of $x$ .

The expression for potential energy of a spring is $U=\frac{1}{2}k{x}^{2}$ .

Given:

Force constant of spring $K=775$ N/m

When $x=\ -5.00$ cm

$x=\ -0.05$ m

The potential energy $K{\ }_{1}=\frac{1}{2}\times{}775\times{}(-0.05{)}^{2}$ J

$K{\ }_{1}=0.96$ J

When $x=\ -4.00$ cm

$x=\ -0.04$ m

The potential energy $K{\ }_{2}=\frac{1}{2}\times{}775\times{}(-0.04{)}^{2}$ J

$K{\ }_{2}=0.62$ J

When $x=\ -3.00$ cm

$x=\ -0.03$ m

The potential energy $K{\ }_{3}=\frac{1}{2}\times{}775\times{}(-0.03{)}^{2}$ J

$K{\ }_{3}=0.34$ J

When $x=\ -2.00$ cm

$x=\ -0.02$ m

The potential energy $K{\ }_{4}=\frac{1}{2}\times{}775\times{}(-0.02{)}^{2}$ J

$K{\ }_{4}=0.155$ J

When $x=\ -1.00$ cm

$x=\ -0.01$ m

The potential energy $K{\ }_{5}=\frac{1}{2}\times{}775\times{}(-0.01{)}^{2}$ J

$K{\ }_{5}=0.038$ J

When $x=0$

$K{\ }_{6}=0$

When $x=1.00$ cm

$K{\ }_{7}=0.038$ J

When $x=2.00$ cm

$K{\ }_{8}=0.155$ J

When $x=3.00$ cm

$K{\ }_{9}=0.34$ J

When $x=4.00$ cm

$K{\ }_{10}=0.62$ J

When $x=5.00$ cm

$K{\ }_{11}=0.96$ J

The potential energy vs x plot is shown as below.

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