Verified step-by-step solutions for your class notes.
Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
The frictional force exerted by the floor is exactly equal to F, acting in the opposite direction. Since the table is moving at a completely constant velocity, its acceleration is zero. This means the net force must be zero, so your forward push and the floor's backward friction completely balance each other out.
For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.
3. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity.
Fig. 6.36a
Fig. 6.36b
Which of the following statement is correct?
The correct choice is: (i) P experiences a net force and Q does not experience a net force.
Block P has unequal forces pulling it in opposite directions (5 N vs 4 N), leaving it with a net force of 1 N. Block Q is moving at a perfect, constant velocity, which always means its acceleration is zero and its net force is absolutely zero.
While practising for the snake boat race, 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat?
The net force on the snake boat is 18,000 N in the forward direction.
Think of it like a giant tug-of-war on the boat: 95 rowers are pulling the right way, generating 19,000 N of forward force (95 × 200 N). Meanwhile, the 5 mistaken rowers are fighting them with 1,000 N of backward force (5 × 200 N). Subtracting the mistake gives a net forward push of 18,000 N.
When a net force acts on an object, we observe that the object accelerates:
The correct choice is (iv) in the direction of force, with acceleration proportional to the force acting on the object. According to Newton's 2nd law, objects always speed up or change direction in the exact path of the net push, and pushing harder directly creates a bigger acceleration.
6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
Object A
Object B
Object C
Object D
A net force acts on: (iii) Object C
Object C's position-time graph is a curved line, which means its speed is changing over time—it is accelerating! In contrast, objects A, B, and D all show completely straight lines, meaning they are either moving at a single constant speed or standing perfectly still with zero net force.
7. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why.
Yes, the boat will move backward (away from the shore). This is a classic example of Newton's 3rd law (Action and Reaction). To launch forward onto the shore, the sailor's feet must push backward against the boat. As the boat pushes the sailor forward, it receives an equal and opposite force that kicks it back into the water.
8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.
The landing mat is there to increase the time it takes for the athlete to come to a complete stop. Sinking into a soft cushion slows down the deceleration process. According to Newton's 2nd law, stretching out that stopping time significantly lowers the impact force hitting the athlete's body, preventing serious injuries.
A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision, which cart exerts a larger force?
The correct option is (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other. Newton's 3rd law states that forces always occur in perfectly equal and opposite pairs. No matter which cart is heavier or moving faster, the actual collision force they trade is exactly identical on both sides.
10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.
The force-mass graph will simply be a flat, horizontal straight line at F = 10 N.
If you take any mass coordinate from the curve and multiply it by its acceleration value (F = m × a), you get the exact same constant force. For instance, at 1 kg mass, acceleration is 10 m/s², which equals 10 N. Since the force remains a constant 10 N for every single mass, its graph is a perfectly flat line at the 10 N mark.
11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
The net force acting on the object is 25 N.
First, find the acceleration by calculating the slope of the line: (30 m/s − 10 m/s) / 8 s = 20 / 8 = 2.5 m/s². Then, plug it into Newton's second law (F = m × a): 10 kg × 2.5 m/s² gives you a total net force of 25 N.
A bullet of mass 50 g moving with a speed of 100 m s–1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet.
The estimated stopping force acting on the bullet is 500 N.
Convert things to standard units first: mass = 0.05 kg (50g) and distance = 0.5 m (50cm). Using the motion equation v² = u² + 2as, we calculate: 0 = 100² + 2 × a × 0.5, giving an acceleration of -10,000 m/s². Finally, using F = ma, the force is 0.05 kg × -10,000 m/s² = -500 N (the negative sign simply shows it opposes the motion).
An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h–1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
The contact time between the foot and the ball is 0.015 seconds.
First, convert the speed from km/h to m/s: 108 × (5/18) = 30 m/s. Since Force = mass × (change in velocity / time), we can set up the calculation: 800 N = 0.4 kg × (30 m/s / t). This simplifies down to 800 = 12 / t, which leaves us with a brief contact window of t = 0.015 seconds.
An object of mass 2 kg moving with a constant velocity of 10 m s–1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied. How much distance does it travel before stopping?
The object travels a distance of 10 meters before coming to a complete stop.
The total combined force fighting the motion is 10 N (7 N rough patch friction + 3 N additional opposition force). With a mass of 2 kg, the deceleration rate is -5 m/s² (Force/mass). Using the formula v² - u² = 2as, we write out 0 - 10² = 2 × (-5) × s, which simplifies cleanly to -100 = -10s, giving a total distance of 10 meters.
A tractor pulls a harrow of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), obtain an expression for the resulting acceleration.
The final mathematical expression for the combined acceleration is: a = (a × a ) / (a + a )
From Newton's second law, mass equals Force divided by acceleration (m = F/a). So, the individual sizes are m = F/a and m = F/a . Combined together, the system mass is (F/a + F/a ). The final combined acceleration is Force divided by this total mass. The 'F' cancels out cleanly from the numerator and denominator, leaving you with the standard product-over-sum formula.
16. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
Even though the magnetic forces acting on both objects are perfectly identical in strength, the bar magnet has vastly more mass than the lightweight compass needle. According to Newton's 2nd law, acceleration is inversely tied to mass. Because the bar magnet is heavy and has static table friction anchoring it down, its movement is completely microscopic, while the tiny needle spins freely.
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