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David Z
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Why What does it mean to say work is path independent-independent when pushing an object in different directions?

Case 1

If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m)$$

Case 2

But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m) $$

  1. If I apply a straight upward(perpendicular to ground) force against gravity of $5\ \mathrm{N}$ and lift an object "A" 10 meters, then the work done is:

    $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J}$$

  2. But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:

    $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J} $$

In case No.cases 1 and 2, the work done is equal, but thethe height would be different in both cases because in case one, height would beis equal to displacement, but in case 2 it will be less than 10 mmeters obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to. Doesn't conflict with the equation work = energy?

P.S Work is force times displacement ($W = F S \cos(x)$$W = F S \cos(\theta)$); where $x$$\theta$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

Why work is path independent?

Case 1

If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m)$$

Case 2

But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m) $$

In case No. 1 and 2 work done is equal, but the height would be different in both cases because in case one height would be equal to displacement, but in case 2 it will be less than 10 m obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to the equation work = energy?

P.S Work is force times displacement ($W = F S \cos(x)$); where $x$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

What does it mean to say work is path-independent when pushing an object in different directions?

  1. If I apply a straight upward(perpendicular to ground) force against gravity of $5\ \mathrm{N}$ and lift an object "A" 10 meters, then the work done is:

    $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J}$$

  2. But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:

    $$ W = F \times S = 5\ \mathrm{N} \times 10\ \mathrm{m} = 50\ \mathrm{J} $$

In cases 1 and 2, the work done is equal, but the height would be different because in case one, height is equal to displacement, but in case 2 it will be less than 10 meters obviously. Then potential energy (i.e mgh) of both objects would be different. Doesn't conflict with the equation work = energy?

Work is force times displacement ($W = F S \cos(\theta)$); where $\theta$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

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Martin
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Case 1

If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is  : -

W = F x S = 5N X10m = 50 J(N-m)$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m)$$

Case 2

But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again  : -

Work = F x S = 5N x 10 m = 50 J(N-m)$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m) $$

In case No. 1 and 2 work done is equal, but the height would be different in both cases because in case one height would be equal to displacement, but in case 2 it will be less than 10 m obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to the equation work = energy?

P.S Work is force times displacement (W = F.S Cosx ;$W = F S \cos(x)$); where x$x$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

Case 1

If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is  : -

W = F x S = 5N X10m = 50 J(N-m)

Case 2

But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again  : -

Work = F x S = 5N x 10 m = 50 J(N-m)

In case No. 1 and 2 work done is equal, but the height would be different in both cases because in case one height would be equal to displacement, but in case 2 it will be less than 10 m obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to the equation work = energy?

P.S Work is force times displacement (W = F.S Cosx ; where x is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

Case 1

If I apply a straight upward(perpendicular to ground) force against gravity of 5N and lift an object "A" 10 meters, then work done is:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m)$$

Case 2

But if I apply the same amount of force diagonally to the ground, and again push the object "A" 10 meters in the direction of force, then again:

$$ W = F \times S = 5\,N \times 10\,m = 50\,J(N-m) $$

In case No. 1 and 2 work done is equal, but the height would be different in both cases because in case one height would be equal to displacement, but in case 2 it will be less than 10 m obviously. Then potential energy (i.e mgh) of both objects would be different, isn't it counter intuitive to the equation work = energy?

P.S Work is force times displacement ($W = F S \cos(x)$); where $x$ is angle between direction of force and displacement) and is path independent, according to most of the text books I've read so far.

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