The required work done to push the steel table is 680.316 J.
Given data:
The mass of block is, m = 13 kg.
The speed of steel table is, v = 1.0 m/s.
The time interval is, t = 8.9 s.
The coefficient of kinetic friction is, [tex]\mu =0.60[/tex].
This problem utilizes the concept of work done due to frictional force. Work done due to frictional force is equal to product of frictional force and distance covered.
The work done due to push the block is given as,
[tex]W = f \times s[/tex] ...................................................(1)
Here, f is the frictional force and s is the distance covered .
And its value is, [tex]s = v \times t[/tex].
Solving as,
[tex]W = (\mu \times m \times g) \times ( v \times t)\\\\W = (0.60 \times 13 \times 9.8) \times ( 1 \times 8.9)\\\\W=680.316 \;\rm J[/tex]
Thus, the required work done to push the steel table is 680.316 J.
Learn more about the work done by frictional force here:
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