A floating ice block is pushed through a displacement d=(15m)i-(12m)j along a straight embankment by rushing water, which exerts a force F=(210N)i-(150N)j on the block. How much work does theforce do on the block during the displacement?

Question

Answer 1

The work done W = F.d

= (210 i - 150 j) \cdot (15 i - 12 j)

= 3150 i - 1800 j

Thus, magnitude of the work done

W = {(F_{x}^{2} + F_{y}^{2})}^{\frac{1}{2}}

Or

W = 3.63 \cdot 10^{3}

J

Answer 2

Solution

We find the work done by the water on the ice block:

$W = \vec{F} \cdot \vec{d}$

$= [(210 N) \hat{i} - (150 N) \hat{j}] \cdot [(15 m) \hat{i} - (12 m) \hat{j}]$

=(210 N)(15 m)+(-150 N)(-12 m)

$= 5.0 \times 10^{3} J$

A floating ice block is pushed through a displacement d=(15m)i-(12m)j along a straight embankment by rushing water, which exerts a force F=(210N)i-(150N)j on the block. How much work does theforce do on the block during the displacement?

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