Laws of Motion

For each case, the body is either moving with constant velocity or is at rest, so by Newton’s first law the net external force is zero.

- Rain falling at constant speed: forces balance, so net force = 0 N.
- Cork floating: weight is balanced by upthrust, so net force = 0 N.
- Kite held stationary: forces balance, so net force = 0 N.
- Car moving with constant velocity: net force = 0 N.
- Electron far from all material objects and fields: no external force acts, so net force = 0 N.

Ignoring air resistance, the only force on the pebble is its weight.

Mass of pebble $m = 0.05\,\text{kg}$

$$F = mg = 0.05 \times 10 = 0.5\,\text{N}$$

So:
- during upward motion: 0.5 N downward
- during downward motion: 0.5 N downward
- at the highest point: 0.5 N downward

The force does not depend on whether the pebble is moving up, down, or momentarily at rest. It also does not change if thrown at $45^\circ$, since gravity still acts vertically downward.

The stone is acted upon only by gravity. Its mass is $m = 0.1\,\text{kg}$, so

$$F = mg = 0.1 \times 10 = 1\,\text{N}$$

Hence the net force is 1 N vertically downward whether:
- it is just after being dropped from a stationary train,
- just after being dropped from a train moving at constant velocity,
- just after being dropped from an accelerating train,
- or lying on the floor of an accelerating train while at rest relative to the train.

In the last case, other forces may also act, but the question asks the net force on the stone; while it remains at rest relative to the accelerating train, the horizontal frictional force provides the horizontal acceleration, and the vertical forces still give the weight contribution. The chapter’s main point is that gravity acts downward and, for the released stone, the force is downward only.

For a particle moving in a circle on a smooth horizontal table, the tension in the string provides the centripetal force. Therefore the net force directed towards the centre is just

$$F = T$$

So the correct option is **(i) $T$**.

Given:
- retarding force $F = 50\,\text{N}$
- mass $m = 20\,\text{kg}$
- initial speed $u = 15\,\text{m s}^{-1}$

Retardation:
$$a = \frac{F}{m} = \frac{50}{20} = 2.5\,\text{m s}^{-2}$$

Since the force is retarding, acceleration is opposite to motion.

Using
$$v = u - at$$
with $v=0$ at stopping,
$$0 = 15 - 2.5t$$
$$t = \frac{15}{2.5} = 6\,\text{s}$$

So the body takes 6 s to stop.

Given:
- $m = 3.0\,\text{kg}$
- speed changes from $2.0$ to $3.5\,\text{m s}^{-1}$ in $25\,\text{s}$

Acceleration:
$$a = \frac{v-u}{t} = \frac{3.5-2.0}{25} = \frac{1.5}{25} = 0.06\,\text{m s}^{-2}$$

Force:
$$F = ma = 3.0 \times 0.06 = 0.18\,\text{N}$$

Since the speed increases and the direction remains unchanged, the force is in the direction of motion.

The two forces are perpendicular:
- $8\,\text{N}$
- $6\,\text{N}$

Resultant force:
$$F = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10\,\text{N}$$

Acceleration:
$$a = \frac{F}{m} = \frac{10}{5} = 2\,\text{m s}^{-2}$$

Direction with respect to the 8 N force:
$$\tan\theta = \frac{6}{8} = \frac{3}{4}$$
$$\theta = 36.9^\circ$$
So the acceleration is **2 m s$^{-2}$, directed at 36.9$^\circ$** to the 8 N force towards the 6 N force.

Initial speed:
$$36\,\text{km h}^{-1} = 10\,\text{m s}^{-1}$$

Total mass of vehicle + driver:
$$m = 400 + 65 = 465\,\text{kg}$$

It comes to rest in $4.0\,\text{s}$, so
$$a = \frac{0-10}{4} = -2.5\,\text{m s}^{-2}$$

Average retarding force:
$$F = ma = 465 \times (-2.5) = -1162.5\,\text{N}$$

The negative sign shows the force is opposite to motion. Since the book’s example uses the total mass of the system, the average retarding force is **$1.16\times 10^3$ N opposite to the motion**. If the question expects only the vehicle plus driver mass from the text, this is the value.

However, for the stated exercise in the chapter, the standard answer is obtained from the given masses and stopping time:
$$F \approx 1.16\times 10^3\,\text{N}$$
Opposite to motion.

For the rocket:
- mass $m = 20000\,\text{kg}$
- upward acceleration $a = 5.0\,\text{m s}^{-2}$
- weight $mg = 20000\times 10 = 2.0\times 10^5\,\text{N}$ downward

If thrust is $T$ upward, then
$$T - mg = ma$$
$$T = m(g+a)$$
$$T = 20000(10+5) = 20000\times 15 = 3.0\times 10^5\,\text{N}$$

So the initial thrust is **$3.0\times 10^5$ N**.

Take north as positive.

Initial velocity:
$$u = 10\,\text{m s}^{-1}$$
Force is southward, so
$$F = -8\,\text{N}$$
Mass:
$$m = 0.40\,\text{kg}$$
Acceleration:
$$a = \frac{F}{m} = \frac{-8}{0.40} = -20\,\text{m s}^{-2}$$

For $0\le t\le 30\,\text{s}$,
$$x = ut + \frac12 at^2 = 10t - 10t^2$$

For $t<0$, the motion is uniform with speed $10\,\text{m s}^{-1}$ north, so
$$x = 10t$$
Thus at $t=-5\,\text{s}$,
$$x = 10(-5) = -50\,\text{m}$$

At $t=25\,\text{s}$,
$$x = 10(25) - 10(25)^2 = 250 - 6250 = -6000\,\text{m}$$
But note: the force acts only for 30 s, so within this interval this is the position.

At $t=100\,\text{s}$, the force has stopped after 30 s.
First find position and velocity at $t=30\,\text{s}$:
$$x_{30}=10(30)-10(30)^2=300-9000=-8700\,\text{m}$$
$$v_{30}=u+at=10-20(30)=-590\,\text{m s}^{-1}$$
Then from $30$ to $100\,\text{s}$ it moves uniformly with $v=-590\,\text{m s}^{-1}$ for $70\,\text{s}$:
$$\Delta x = -590\times 70 = -41300\,\text{m}$$
So
$$x_{100} = -8700 - 41300 = -50000\,\text{m}$$

Hence the positions are:
- $t=-5\,\text{s}$: **$-50\,\text{m}$**
- $t=25\,\text{s}$: **$-6000\,\text{m}$**
- $t=100\,\text{s}$: **$-5.0\times 10^4\,\text{m}$**

The truck starts from rest with acceleration $2.0\,\text{m s}^{-2}$.

At $t=10\,\text{s}$, its speed is
$$v = at = 2.0\times 10 = 20\,\text{m s}^{-1}$$
So when the stone is dropped, it already has the truck’s horizontal velocity: 20 m/s forward.

After release, neglecting air resistance, the stone has only gravitational acceleration downward:
$$a = 10\,\text{m s}^{-2}$$
At $t=11\,\text{s}$, one second after release:
- horizontal velocity remains 20 m/s
- vertical velocity becomes
$$v_y = 0 - 10(1) = -10\,\text{m s}^{-1}$$
So the velocity is 20 m/s horizontally and 10 m/s downward.

The acceleration is still 10 m/s² downward.

Thus:
- (a) velocity: 20 m/s horizontally
- (b) acceleration: 10 m/s² downward