Question

g The force that a 60 Kg block applies to an ideal spring (initially in its equilibrium position) produces a work 90 J , and stretches the spring 0.1 cm . How much work is required to stretch the spring by 1 cm from its equilibrium position

Answer 1

Answer:

Work = 9000J

Explanation:

Work done on the spring is 60J

mass of block 60kg

extension on the spring 0.1cm

Workdone = $\frac{1}{2}$Ke²

where K is the force constant of the spring, Inputting the values

we have 90 = $\frac{1}{2}$×K×0.1²

K = $\frac{180}{0.01}$

K = 18000 N/cm

How much work is required to stretch the spring by 1 cm from its equilibrium position, we have

Workdone = $\frac{1}{2}$Ke²

work = $\frac{1}{2}$×18000×1²

Work = 9000J

= 9KJ

Answer 2

Answer:

W = 9 kJ

Explanation:

We have:

m: mass of the block = 60 kg

W: work = 90 J

x: distance = 0.1 cm = 1.00x10⁻³ m                

To find the work we can use the following equation:

$W = F*x = \frac{1}{2}kx^{2}$                                              

Where:

F: is the force that the block applies to an ideal spring = -kx    

First, we need to find the spring constant:

$k = \frac{2W}{x^{2}} = \frac{2*90 J}{(1.00 \cdot 10^{-3} m)^{2}} = 1.80 \cdot 10^{8} N/m$  

Now, with the spring constant we can find the work required to stretch the spring by 1 cm from its equilibrium position:

$W = \frac{1}{2}kx^{2} = \frac{1}{2}1.80 \cdot 10^{8} N/m*(1.00 \cdot 10^{-2} m)^{2} = 9000 J = 9 kJ$    

Therefore, the work required is 9 kJ.

I hope it helps you!