Physics (Work done by a varying force)

Question

Five kilograms of mangoes is bought on a market and measured using a vertical spring scale that obeys Hookes’s law, it stretches up to 3.00 cm. The mangoes were removed and replaced by kiwi

a. If the scale shows 8.00 kg, how far will the spring stretch? (Ans. 0.048 m)

b. How much work must the external agent do to stretch the spring to 8.00 cm from its unstretched position. (Ans. 5.23 J)

Please show solution, I've been getting a different value from the given answers.

Answer 1

Five kilograms of mangoes is bought on a market and measured using a vertical spring scale that obeys Hookes’s law, it stretches up to 3.00 cm. The mangoes were removed and replaced by kiwi

a. If the scale shows 8.00 kg, how far will the spring stretch? (Ans. 0.048 m)

b. How much work must the external agent do to stretch the spring to 8.00 cm from its unstretched position. (Ans. 5.23 J)

Please show solution, I've been getting a different value from the given answers.

Hello Jacob

Let us first find the spring constant

k= Force/extension= 5g/3*10^-2 N/m

Next time

Extension = Force /k= 8g/k = (8g* 3*10^-2 )/5g) = 24*10^-2 / 5 = =4.8*10^-2 m = 0.048 m

Work done = 1/2 ( k x² )

where k = spring constant and x = extension

Plug in k and x= 8*10^-2 m and get energy in joules

Answer 2

Hooke's Law:

F = k x and

Elastic potential energy

U = 1/2 k x²

Draw a diagram for yourself. If you do not know how to, check out any text or online source.

From the information given about the mangoes in the question, We use the following to find the k first.

F = mg = k x where m is the mass, g gravitational acceleration, we use 10 m/s² here for simplicity!

put in the numbers

(5.00 Kg)(10 m/s²) = k (3.00 x 10^-2 m) assuming 3 S.F. for mass.

This gives: k = (5.00 Kg)(10 m/s²) / (3.00 x 10^-2 m)

cancel the units, we have:

k = (5.00 Kg)(10 m/s²) / (3.00 x 10^-2 m)

= 1.67 x 10³ Kg/s²

Now, for kiwi, it is 8.00 Kg, so the x is:

F = mg = k x = 8.00 Kg x 10 m/s² = (1.67 x 10³ Kg/s²) • x

solve for x:

x = (8.00 Kg x 10 m/s²) / (1.67 x 10³ Kg/s²)

Cancel the units resulting the m only:

x = (8.00 Kg x 10 m/s²) / (1.67 x 10³ Kg/s²)

= (8.00 x 10) / (1.67 x 10³) m

= 8.00 / (1.67 x 10²) m

= 4.79 x 10^-2 m // answer OR 4.79 cm //

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Work done to the spring equals to the change of elastic potential energy of the spring, such that

W = ΔU = U_f - U_i

W = 1/2 k (x_f² - x_i²) = ΔU = U_f - U_i

Set Ui to zero joule when x = 0.00 cm in equilibrium position, then we have:

W = 1/2 k x_f² = U_f = ΔU

if now we have x_f = 8.00 cm, so:

W = 1/2 x (1.67 x 10³ Kg/s²) x (8.00 x 10^-2 m)²

= 5.34 Kg•m²/s²

= 5.34 J // answer

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It will be important to use units during the calculation to ensure the physical meanings are followed throughout the process. Besides, you learn the dimensional analysis skills from the calculation process. When you are comfortable with your understanding of the many physics principles, then you can see through every calculation you or others made!!