1 answer

A 2.0 kg block is sliding across a table with an initial speed of 0.5 m/s....

Question:

A 2.0 kg block is sliding across a table with an initial speed of 0.5 m/s. The table imparts a 20.0 N force due to friction on the block. How much work in Joules will the table do on the block?


Answers

Work done by friction force will be equal to change in kinetic energy of block.

Wfriction = Kf - Ki​​​​​​

  = 0 - mu2/2

= 0 - 2x(0.52)/2

Wfriction = - 0.25 Joule

.

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