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NCERT Solutions · Class 9 Science How Forces Affect Motion

26 questions · 19 still being checked

Pause and Ponder 6.1–6.10 (part 1 of 3)

  1. Exercise 6.1

    A weightlifter lifts a barbell (Fig. 6.8\displaystyle 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?NCERT_Question_Class9_Science_Ch6_PP_Q6-1

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    The two forces on the barbell are the gravitational force (its weight), acting downwards, and the upward force applied by the weightlifter's hands.
    Yes — if the barbell is held steady, these two forces are balanced.
    Steady means the barbell is at rest: its velocity is zero and unchanging, so by Newton's first law the net force on it must be zero.
    Two forces balance when they are equal in magnitude and opposite in direction, so the lifter's upward push exactly equals the weight.
    For the barbell of Example $\displaystyle 6.4$ (total mass $\displaystyle 30$ kg), the weight is \(\displaystyle F = mg = 30\ \text{kg} \times 9.8\ \text{m s}^{-2} = 294\ \text{N} \) downwards, so the lifter applies \(\displaystyle 294\ \text{N} \) upwards.

  2. Exercise 6.2

    Two players R and S are participating in an arm-wrestling match (Fig. 6.9\displaystyle 6.9). At the instant, when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?NCERT_Question_Class9_Science_Ch6_PP_Q6-2

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    NCERT’s answer
    No; S
    No, the forces are not balanced — player S exerted the larger force.
    If the two players pushed with equal and opposite forces, the net force on the joined arms would be zero and the arms would stay put.
    The arms do move, so a non-zero net force acts on them.
    When two opposite forces are unequal, the net force equals the difference of their magnitudes and points along the larger force.
    The arms tilt out of the page towards you, so the larger force is the one pushing in that direction — in Fig. $\displaystyle 6.9$ that is S.

  3. Exercise 6.3

    An object is moving with a constant velocity. Is there a net force acting upon it?

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    NCERT’s answer
    No
    No — no net force acts on it.
    Constant velocity means the magnitude and direction of the velocity are both unchanging, so the acceleration is zero.
    Newton's first law: an object in motion continues to move with a constant velocity unless a net force acts upon it.
    Individual forces may well be acting (weight, normal force, friction, an applied push), but they cancel out, so the net force is \(\displaystyle 0\ \text{N} \).

  4. Exercise 6.4

    Suppose, no net force is acting on an object. Which of the following situations are possible? (i) Object remains at rest if at rest. (ii) Object keeps moving with a constant velocity if already moving. (iii) Object is moving with a constant acceleration.

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    NCERT’s answer
    (i)
    ; (ii)
    (i) and (ii) are possible; (iii) is not.
    (i) Possible — with zero net force an object at rest stays at rest (Newton's first law).
    (ii) Possible — an object already moving keeps moving with the same constant velocity.
    (iii) Not possible — constant acceleration \(\displaystyle a \) needs a net force \(\displaystyle F = ma \), and with \(\displaystyle F = 0\ \text{N} \) we get \(\displaystyle a = 0\ \text{m s}^{-2} \), so the object cannot be accelerating.

  5. Exercise 6.5

    In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.

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    Example: pushing a box across the floor so that it moves with a constant velocity.
    Four forces act on the box — the applied push forwards, the force of friction backwards, the weight downwards and the normal force from the floor upwards.
    The weight and the normal force are already balanced (Fig. $\displaystyle 6.11$).
    If you now push with a force exactly equal in magnitude to the force of friction, those two also balance, so the net force becomes zero even though four forces are acting.
    The box then keeps moving with constant velocity, as in Example 6.2.
    Another everyday case: a weightlifter holding a barbell steady — the upward push she adds makes the net force on the barbell zero.
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-5

  6. Exercise 6.6

    A toy car of mass 100\displaystyle 100 g is moving with a constant velocity of 0.5\displaystyle 0.5 m s−1\displaystyle s^{-1}. What is the net force acting on the toy car?
    NCERT’s answer
    $\displaystyle 0$ N
    The net force is zero, \(\displaystyle 0\ \text{N} \).
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-6
    Mass \(\displaystyle m = 100\ \text{g} = 0.1\ \text{kg} \), and the velocity is constant at \(\displaystyle 0.5\ \text{m s}^{-1} \).
    Constant velocity means no change in velocity, so \(\displaystyle a = 0\ \text{m s}^{-2} \).
    By Newton's second law, \(\displaystyle F = ma = 0.1\ \text{kg} \times 0\ \text{m s}^{-2} = 0\ \text{N} \).

  7. Exercise 6.7

    Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

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    A larger force is needed for the heavier child (the one of greater mass).
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-7
    Newton's second law gives \(\displaystyle F = ma \), so for the same acceleration \(\displaystyle a \) the force needed is directly proportional to the mass.
    If one child has mass \(\displaystyle m_1 \) and the other \(\displaystyle m_2 \) with \(\displaystyle m_2 > m_1 \), then \(\displaystyle F_2 = m_2 a > m_1 a = F_1 \).
    For example, to give an acceleration of \(\displaystyle 2\ \text{m s}^{-2} \) a $\displaystyle 20$ kg child needs \(\displaystyle 20 \times 2 = 40\ \text{N} \), while a $\displaystyle 40$ kg child needs \(\displaystyle 40 \times 2 = 80\ \text{N} \).

  8. Exercise 6.8

    How are glass items packed for transportation using a bubble wrap or hay protected from damage?

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    Bubble wrap and hay protect glass by increasing the time over which a jolt changes the item's velocity, which reduces the force on it.
    During transport a glass item that is jolted or dropped has to be brought to rest.
    A soft, compressible packing squashes as the item pushes into it, so the velocity falls to zero over a longer time interval instead of instantly.
    A longer stopping time means a smaller magnitude of acceleration, and by \(\displaystyle F = ma \) a smaller force acts on the glass.
    Hitting a hard crate wall directly would stop the item in a very short time, needing a very large force — enough to crack it, just as the ground's large force cracks a coconut (Fig. $\displaystyle 6.20$).
    This is the same principle as a cricketer drawing their hands back while catching, and as a vehicle airbag.
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-8

  9. Exercise 6.9

    Why does a fireperson sometimes struggle when holding the pipe issuing water?

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    Because of Newton's third law — the pipe pushes the water forward, so the water pushes the pipe backward with an equal force.
    The hose forces a large amount of water out of the nozzle at high velocity, i.e. it exerts a large forward force on the water.
    Simultaneously the water exerts an equal and opposite force on the pipe, directed backwards.
    This backward (recoil) force tends to push the pipe out of the fireperson's hands, so they must apply a large force to hold it steady — which is a struggle.
    It is the same effect that drives the balloon in Activity $\displaystyle 6.7$ and lifts a rocket (Fig. $\displaystyle 6.30$).
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-9

  10. Exercise 6.10

    Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how can it change its velocity.

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    It can change its velocity by firing its engine to expel gas, using Newton's third law.
    NCERT_Solution_Class9_Science_Ch6_PP_Q6-10
    The engine pushes exhaust gas out in one direction with a large force.
    The gas simultaneously pushes the spacecraft with an equal force in the opposite direction.
    This is an external force on the spacecraft, so by \(\displaystyle F = ma \) it produces an acceleration and the velocity changes.
    Firing the gas backwards speeds the craft up; firing it in the direction of motion slows it down — exactly what the Vikram lander of Chandrayaan-$\displaystyle 3$ did to slow for its soft landing near the Moon's south pole.
    Note that no gravitational force or contact with any surface is needed: the craft pushes on its own exhaust.