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

Work = 9000J

Explanation:

Work done on the spring is 60J

mass of block 60kg

extension on the spring 0.1cm

Workdone = Ke²

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

we have 90 = ×K×0.1²

K =

K = 18000 N/cm

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

Workdone = Ke²

work = ×18000×1²

Work = 9000J

= 9KJ

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:

Where:

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

First, we need to find the spring constant:

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

Therefore, the work required is 9 kJ.

I hope it helps you!

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

answersysvol bloat;