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Exercises · 4.6

Q.A constant force acting on a body of mass 3.0 kg3.0\ \text{kg} changes its speed from 2.0 m s−12.0\ \text{m s}^{-1} to 3.5 m s−13.5\ \text{m s}^{-1} in 25 s25\ \text{s}. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

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A constant force changes the velocity of a body, so we find the acceleration from the velocity change and time, then apply Newton's second law F=maF = ma to get the force. The magnitude is 0.18 N\boxed{0.18\ \text{N}} in the direction of motion.

When a constant force acts on a body, it produces a constant acceleration. Newton's second law tells us that force and acceleration are directly proportional: F=maF = ma. The strategy is straightforward—find the acceleration from the change in velocity, then multiply by mass to get the force.

Since the direction of motion remains unchanged throughout, this is one-dimensional motion. The body is speeding up (from 2.0 m s−12.0\ \text{m s}^{-1} to 3.5 m s−13.5\ \text{m s}^{-1}), so the force must act in the same direction as the velocity.

Finding the force

  1. Calculate the acceleration

    Acceleration is the rate of change of velocity. With initial speed u=2.0 m s−1u = 2.0\ \text{m s}^{-1}, final speed v=3.5 m s−1v = 3.5\ \text{m s}^{-1}, and time interval t=25 st = 25\ \text{s}:

a=v−ut=3.5−2.025=1.525=0.06 m s−2a = \frac{v - u}{t} = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06\ \text{m s}^{-2}

The positive value confirms the acceleration is in the direction of motion (the body is speeding up).

  1. Apply Newton's second law

    With mass m=3.0 kgm = 3.0\ \text{kg} and acceleration a=0.06 m s−2a = 0.06\ \text{m s}^{-2}: …

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