Chapter 10, Problem 39A

To determine

Watt equivalent in terms of kilograms, meters and seconds.

Answer to Problem 39A

Watt equivalent in terms of kilograms, meters and seconds is $\text{W}=\frac{{\text{kg-m}}^{\text{2}}}{{\text{s}}^{\text{3}}}$

Explanation of Solution

Power can be expressed as:

As, $P=\frac{W}{t}$

Where, W is work done which is measured in Joules and t is the time measured in seconds.

i.e.

  $\begin{array}{l}\text{W}=\frac{\text{J}}{\text{s}}=\frac{\text{N-m}}{\text{s}}=\text{N}\frac{\text{m}}{\text{s}}\\ =\frac{\text{kg-m}}{{\text{s}}^{\text{2}}}\times \frac{\text{m}}{\text{s}}\\ =\frac{{\text{kg-m}}^{\text{2}}}{{\text{s}}^{\text{3}}}\end{array}$