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3 years ago

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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