A body of mass $0.40\ \text{kg}$ moving initially with a constant speed of $10\ \text{m s}^{-1}$ to the north is subject to a constant force of $8.0\ \text{N}$ directed towards the south for $30\ \text{s}$. Take the instant the force is applied to be $t = 0$, the position of the body at that time to be $x = 0$, and predict its position at $t = -5\ \text{s},\ 25\ \text{s},\ 100\ \text{s}$.

A body of mass 0.40kg moving initially with a constant speed of 10m s⁻¹ to the north is subject to a constant

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0.40\ \text{kg}

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Concept understanding — Newton Second Law

Newton's Second Law: The Law That Connects Force and Motion

Imagine you're pushing a shopping cart. If you push gently, it moves slowly. Push harder, and it speeds up faster. Now imagine the cart is full of groceries — even with the same push, it accelerates much more slowly than an empty cart. This everyday experience is exactly what Newton's Second Law captures.

The Intuition First

Two things matter when you push something:

How hard you push — the force you apply.
How heavy the object is — its mass.

The harder you push, the more the object speeds up. The heavier the object, the less it speeds up for the same push. So acceleration depends on both force and mass — and in opposite ways.

Note

"Acceleration" here means any change in velocity — speeding up, slowing down, or changing direction. It's not just "going faster."

The Precise Statement

Newton's Second Law says:

The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The acceleration is in the same direction as the net force.

In one equation:

\vec{a} = \frac{\vec{F}_{\text{net}}}{m}

Or more commonly:

\vec{F}_{\text{net}} = m \vec{a}

Where:

$\vec{F}_{\text{net}}$ is the net force (the vector sum of all forces acting on the object) — measured in newtons (N)
$m$ is the mass of the object — measured in kilograms (kg)
$\vec{a}$ is the acceleration — measured in metres per second squared ( $\text{m/s}^2$ )

\vec{F}_{\text{net}} = m \vec{a}

What This Really Means

Force causes acceleration, not velocity. A constant net force produces constant acceleration — meaning the velocity keeps changing at a steady rate. If you stop pushing, the net force becomes zero, and acceleration becomes zero (the object continues at constant velocity — that's Newton's First Law).

Mass is a measure of inertia. The more mass an object has, the harder it is to change its motion. A truck needs a much larger force than a bicycle to achieve the same acceleration.

Direction matters. Force and acceleration are vectors — they point the same way. If you push north, the acceleration is north. If multiple forces act, you must add them as vectors to find the net force.

A Simple Example

A 2 kg block is pushed with a net force of 10 N to the right.

a = \frac{F_{\text{net}}}{m} = \frac{10\ \text{N}}{2\ \text{kg}} = 5\ \text{m/s}^2

The block accelerates at $5\ \text{m/s}^2$ to the right. Every second, its velocity increases by 5 m/s in that direction. …

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The problem uses Newton's second law to find motion under a constant opposing force. The body moves north at constant speed until $t=0$ , then decelerates southward. Splitting time into before, during, and after the force gives: $x(-5) = -50\ \text{m}$ , $x(25) = -6000\ \text{m}$ , $x(100) = -50000\ \text{m}$ .

The key here is that the force is constant and opposes the initial velocity. Newton's second law tells us the acceleration is constant, so we can use the kinematic equations for uniform acceleration — but only after the force starts. Before $t=0$ , there is no force, so the body moves with constant velocity.

Let's set up the coordinate system: north is positive $x$ , south is negative $x$ .

At $t=0$ , the body is at $x=0$ and moving north at $10\ \text{m/s}$ .

The force of $8.0\ \text{N}$ south means a negative acceleration:

a = \frac{F}{m} = \frac{-8.0}{0.40} = -20\ \text{m/s}^2

That's a large deceleration — the body will slow, stop, and then accelerate southward.

Now we handle each requested time.

At $t = -5\ \text{s}$ — before the force acts. From $t=-5$ to $t=0$ , motion is uniform northward at $10\ \text{m/s}$ . Displacement in that $5\ \text{s}$ interval:

\Delta x = v \cdot \Delta t = 10 \times 5 = 50\ \text{m}

Since at $t=0$ we are at $x=0$ , at $t=-5$ we were $50\ \text{m}$ south of that: $x = -50\ \text{m}$ .

Watch out

A common mistake is to think $x(-5) = +50\ \text{m}$ . But moving north means $x$ increases; if at $t=0$ we are at $0$ , then 5 seconds earlier we were further south (negative $x$ ).

At $t = 25\ \text{s}$ — after the force starts. Use $x = v_0 t + \frac12 a t^2$ with $v_0 = 10\ \text{m/s}$ , $a = -20\ \text{m/s}^2$ , $t=25$ :

x(25) = 10 \times 25 + \frac12 (-20) \times (25)^2

= 250 - 10 \times 625 = 250 - 6250 = -6000\ \text{m}

That's $6000\ \text{m}$ south of the origin. Checking when the body momentarily stops:

$v = v_0 + a t = 0 \Rightarrow 10 - 20 t = 0 \Rightarrow t = 0.5\ \text{s}$ .

So after just half a second, it stops and then moves south. At $t=25$ , it's been moving south for $24.5\ \text{s}$ — hence the large negative position.

At $t = 100\ \text{s}$ — the force acts for only $30\ \text{s}$ , a crucial detail. From $t=0$ to $t=30$ , acceleration is $-20\ \text{m/s}^2$ . After $t=30$ , the force stops, so acceleration becomes zero again — the body moves with constant velocity (the velocity it had at $t=30$ ). …

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