Force Exerted by a Water Jet – Rankers Physics
Topic: Laws of Motion
Subtopic: Definition of Force Inertia and First Law

Force Exerted by a Water Jet

An object of mass 2 kg is placed on a smooth horizontal surface. A water jet throws water at a rate of 2 kg/s with speed 20 m/s which strikes the object horizontally. The force applied by water jet and acceleration of the object respectively, are
\[20 N, 40 m/s^2\]
\[40 N, 20 m/s^2\]
Zero, zero
\[12 N, 12 m/s^2\]

Solution:

Formula: \(F = v \frac{dm}{dt} = 20 \times 2 = 40\text{ N}\). Using Newton's second law, acceleration \(a = \frac{F}{m} = \frac{40}{2} = 20\text{ m/s}^2\).

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