Upon completing this lesson, you will be able to:
In everyday language, "work" can mean many things. In physics, work has a very precise definition: work is done on an object when a force causes a displacement of that object.
For work to be done, two conditions must be met:
The work $(W)$ done by a constant force $(F)$ on an object that undergoes a displacement $(d)$ is given by the product of the magnitude of the displacement times the component of the force parallel to the displacement:
$$ W = \vec{F} \cdot \vec{d}=Fd\cos(\theta) $$
SI units work is measured in newton-meters A special name is given to this unit, the joule ($J$)
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A person pulls a $50$kg crate $40$m along a horizontal floor by a constant force $F_P=100$N which acts at a $37°$ angle. The floor is rough and exerts a friction force, $F_{fr}=50$N Determine the work done by each force acting on the crate, and the net work done on the crate.
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The work done by the gravitational and normal forces is zero, since they are perpendicular to the displacement $\vec{\mathbf{x}}$: $\theta=90\degree$
$W_G = mg\times x \times \cos(90°) = 0$
$W_N = F_N\times x \times \cos(90°) = 0$
The work done by $\vec{\mathbf{F_p}}$ is:
$$ W_P = F_P \times \cos(\theta) = 100 \times 40 \times \cos(37) = 3200 \text{J} $$
The work done by Friction Force is:
$$ W_{fr} = F_{fr} \times \cos(180\degree) = 40 \times 40 \times \cos(180) = -2000 \text{J} $$
Net Work:
$$ W_{net} = W_{G}+W_{N}+W_{P}+W_{fr}=0+0+3200-2000=1200\text{J} $$
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Determine the work a hiker must do on a $15.0$kg backpack to carry it up a hill of height $h=10.0$m. Determine also the work done by gravity on the backpack, and the net work done on the backpack. For simplicity, assume the motion is smooth and at constant velocity (i.e., acceleration is zero).
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The forces on the backpack:
the force of gravity, $m\vec{\mathbf{g}}$, acting downward; and $\vec{\mathbf{F_H}}$ the force the hiker must exert upward to support the backpack. The acceleration is zero, so horizontal forces on the backpack are negligible.
We are interested in the vertical motion of the backpack, so we choose the y coordinate as positive vertically upward.
Newton’s second law applied in the vertical direction to the backpack gives:
$$ \sum F_y=ma_y \\ F_H-mg=0 \\a_y=0 \text{, so} \\ F_H=mg=(15)(9.81)=147 \text{N} $$
To calculate the work done by the hiker on the backpack, we write:
$$ W_H=F_H(d \cos(\theta)) $$
and we note that $d \cos(\theta)=h$. So the work done by the hiker is
$$ W_H=F_H(d \cos(\theta))=F_Hh=mgh \\=(147)(10)=1470 \text{J} $$
Note that the work done depends only on the change in elevation and not on the angle of the hill $\theta$, The hiker would do the same work to lift the pack vertically the same height h.
The work done by gravity on the backpack is:
$$ W_G=F_Gd( \cos(180-\theta)) \\\cos(180-\theta)=-\cos(\theta) \\ \\ W_G=F_Gd(- \cos(\theta))=-mgh \\=-(147)(10)=-1470 \text{J} $$
The work done by gravity (which is negative here) doesn’t depend on the angle of the incline, only on the vertical height $h$ of the hill. This is because gravity acts vertically, so only the vertical component of displacement contributes to work done
Net Work:
$$ W_{net} = W_{G}+W_{H}=1470-1470=0\text{J} $$
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