Physics College

A force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.
A). What is the potential energy of the spring when it is stretched 0.700 m?
B). What is its potential energy when it is compressed 8.00 cm?

Answers

Answer 1

Answer:

399j and 456j

Explanation:

(a)spring energy= force * distance

spring energy=570 * 0.700= 399 joules

(b) convert 8.00 cm to m we have 0.08m

E =f * s

E=570 *0.08= 456j

Answer 2

The potential energy of the spring when it is stretched 0.700 m would be  199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.

What is the spring constant?

The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.

The mathematical relation for calculating the spring constant is as follows

F = - Kx

As given in the problem, if a  force of 570 N keeps a certain ideal spring stretched at a distance of 0.700 m.

k = 570 / 0.7

k = 814.28 N / m

 

The potential energy stored  in spring when it is stretched 0.700 m

E = 1/2 kx²

 =0.5×814.28×0.7²

 = 199.5 Joules

The potential energy stored  in spring when it is stretched 8 cm

E = 1/2 kx²

 =0.5×814.28×0.08²

 =2.60 Joules

Thus, the potential energy of the spring when it is stretched 0.700 m would be  199.5 Joules, and the potential energy when it is compressed 8.00 cm would be 2.60 Joules.

Learn more about spring constants here,

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Related Questions

A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​

Answers

A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

How to calculate  the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​?

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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