A scoeter \( (m=150 / \mathrm{kg}) \) moves at a constant ipeed of \( 30 \mathrm{~ms}^{-1} \) on a straght flat road he power delivered by the scooter is 2400 W What is the frictional fore that eppeses the forward motion?

To find the frictional force that opposes the forward motion of the scooter, we can use the relationship between power, force, and velocity. The power delivered by the scooter is given by the formula: \[ P = F \cdot v \] where: - \( P \) is the power (in watts), - \( F \) is the force (in newtons), - \( v \) is the velocity (in meters per second). Given: - \( P = 2400 \, \text{W} \) - \( v = 30 \, \text{m/s} \) We can rearrange the formula to solve for the frictional force \( F \): \[ F = \frac{P}{v} \] Substituting the given values: \[ F = \frac{2400 \, \text{W}}{30 \, \text{m/s}} = 80 \, \text{N} \] Thus, the frictional force that opposes the forward motion of the scooter is \( 80 \, \text{N} \).