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You want to launch a stone using the elastic band of a slingshot. The force that the elastic band applies to an object is given by F = −α∆s4 , where α = 45 N m4 and ∆s is the displacement of the elastic band from its equilibrium position. You load the slingshot with the stone and pull back on the stone horizontally, stretching the the elastic band 20 cm. How much work do you do on the stone-elastic band system?

Respuesta :

Manetho

Answer:

0.00288 J

Explanation:

We know that

W= Fds

F = −α∆s^4

α = 45 N/m^4 and ∆s = displacement

W= −α∆s^4ds

integrating both the sides from s= 0 to 0.2

W= 45/5×0.2^5= 0.00288 J

okpalawalter8

The workdone on the stone-elastic band system is mathematically given as

W= 0.00288 J

What work do you do on the stone-elastic band system?

Question Parameter(s):

F = −α∆s4 , where α = 45 N m4

∆s is the displacement of the elastic band

Generally, the equation for the Workdone  is mathematically given as

W= Fds

Thereofre

F = −α *ds^4

Where

α = 45  

ds = displacement

In conclusion

W= 45/5*0.2^5

W= 0.00288 J

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