8.2 Newton's second law of motion

The only force acting on the ball is gravity (gravitational force due to Earth). It remains constant as acceleration due to gravity is constant in the vicinity of Earth. Thus, net force on the ball during its flight remains constant, which is equal to the weight of the ball i.e.

$$F=\mathrm{mg}=0.1X10=1N$$

Hence, option (d) is correct.

In the case (a), the pebble is moving with horizontal acceleration of 10 $m/s^2$ as seen from the ground reference (inertial frame of reference). The net external force in horizontal direction is :

$$\Rightarrow F_{\text{net}}=\mathrm{ma}=0.01X10=1N\text{(acting horizontally)}$$

There is no vertical acceleration. As such, there is no net external force in the vertical direction. The magnitude of net force on the pebble is, therefore, 1 N in horizontal direction.

In the case (b), the pebble is moving with uniform velocity of 10 m/s in horizontal direction as seen from the ground reference (inertial frame of reference). There is no net force in horizontal direction. When dropped, only force acting on it is due to gravity. The magnitude of net force is, thus, equal to its weight :

$$\Rightarrow F_{\text{net}}=\mathrm{mg}=0.01X10=1N\text{(acting downward)}$$

In the case (c), the pebble is moving with acceleration of 10 $m/s^2$ as seen from the ground reference (inertial frame of reference). When the pebble is dropped, the pebble is disconnected with the accelerating train. As force has no past or future, there is no net horizontal force on the pebble. Only force acting on pebble is its weight. The magnitude of force on the pebble is :

$$\Rightarrow F_{\text{net}}=\mathrm{mg}=0.01X10=1N\text{(acting downward)}$$

Hence, option (d) correct.

The net force on the rocket is difference of the thrust and weight of the rocket (thrust being greater). Let thrust be F, then applying Newton’s second law of motion :

$$F_{\text{net}}=F-\mathrm{mg}=\mathrm{ma}$$

$$\Rightarrow F=m(g+a)=10000X(10+10)=200000=2X{10}^{5}N$$

Hence, option (b) is correct.

The given equation of displacement is a quadratic equation in time. This means that the object is moving with a constant acceleration. Comparing given equation with the standard equation of motion along a straight line :

$$x=\mathrm{ut}+\frac{1}{2}a{t}^2$$

We have acceleration of the object as :

$$a=5X2=10m/t^2$$

Applying Newton’s second law of motion :

$$F=\mathrm{ma}=1X10=10N$$

Hence, option (d) is correct.

The plank applies resistance as force. This force decelerates the bullet and is in opposite direction to the motion of bullet. Since we seek to know average acceleration, we shall consider this force as constant force assuming that wooden has uniform constitution. Let the corresponding deceleration be “a”. Then, according to equation of motion,

$$v^2=u^2+2\mathrm{as}$$

$$\Rightarrow a=-\frac{v^2-u^2}{2s}$$

Putting values,

$$\Rightarrow a=-\frac{{10}^4-0}{2X0.5}=-{10}^{4}m/t^2$$

Applying Newton’s second law of motion :

$$\Rightarrow F=\mathrm{ma}=0.01X{10}^4=100N$$

Hence, option (b) is correct.