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KatRina [158]
3 years ago
6

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 1/2 mv2

Physics
2 answers:
snow_lady [41]3 years ago
7 0

Explanation :

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by \dfrac{1}{2}mv^2.

We know that work done is equal to the change in kinetic energy.

                     W=\Delta KE

Let kinetic energy increases from 0 to \dfrac{1}{2}mv^2

 So,                F.s=\Delta KE

                       F.s=\dfrac{1}{2}mv^2........................(1)

According to second condition if both force and distance is tripled, then kinetic energy increases by KE^'.

So,                        3F\times 3s=\Delta KE^'

                               9F.s=\Delta KE'..........................(2)

Using equation (1) in (2)

\dfrac{9}{2}mv^2=\Delta KE^'

This is the resulting increase in kinetic energy.

Original increase in kinetic energy is \dfrac{9}{2}mv^2.

agasfer [191]3 years ago
6 0
W work
F force 
s distance

If F = constant:

W₁ = F·s

If you triple the force and the distance:

W₂ = 3F · 3s = 9 F·s = 9 W₁

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