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How Forces Affect Motion

Here are the cleaned solutions with all dollar signs removed for clarity:

1. 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?

• Answer: Since the table is moving at a constant velocity, its acceleration is zero (a=0). According to Newton’s first law of motion, the net force acting on the table must be zero. Therefore, the applied horizontal force F is exactly balanced by the opposing force of friction.
• Magnitude: The frictional force exerted by the floor is equal in magnitude to the applied force, i.e., F, acting in the direction opposite to the motion.

2. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.

• (i) If no net force is applied on the ball, the velocity of the ball will: remain the same. (In accordance with Newton’s first law, absence of a net force means uniform motion continues).
• (ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will: increase.
• (iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will: decrease.

3. Which of the following statements is correct for blocks P and Q?

• Context: Block P has two opposite forces of 4 N and 5 N acting on it, while block Q is moving with a constant velocity.
• Analysis:
• 
  
  • For block P, the net force is 5 N – 4 N = 1 N (non-zero), so P experiences a net force.
  • For block Q, since it moves with a constant velocity, its acceleration is zero, meaning the net force acting on it is zero (Net Force = 0). Thus, Q does not experience a net force.
• Correct Option: (i) P experiences a net force and Q does not experience a net force.

4. What is the net force on the snake boat?

• Given: Total oarsmen = 100.
• 
  
  • Oarsmen rowing backward (propelling forward) = 95.
  • Oarsmen rowing in the opposite direction (mistake) = 5.
  • Force applied by each oarsman = 200 N.
• Calculation:
• 
  
  • Forward force (Fforward) = 95 x 200 N = 19,000 N
  • Backward/Opposing force (Fbackward) = 5 x 200 N = 1,000 N
  • Net force = Fforward – Fbackward = 19,000 N – 1,000 N = 18,000 N
• Answer: The net force on the snake boat is 18,000 N in the forward direction.

5. When a net force acts on an object, we observe that the object accelerates:

• Correct Option: (iv) in the direction of force, with acceleration proportional to the force acting on the object. (This directly reflects Newton’s second law: a is proportional to F in the direction of the net force).

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:

• Analysis of Graphs (Fig. 6.37):
• 
  
  • Object A: Straight-line graph with constant slope Constant velocity Zero acceleration/net force.
  • Object B: Horizontal line Stationary (at rest) Zero acceleration/net force.
  • Object C: Curved upward graph Changing slope (increasing velocity/acceleration) Non-zero net force acts on C.
  • Object D: Straight line sloping downwards Constant negative velocity Zero acceleration/net force.
• Correct Option: (iii) Object C

7. A sailor jumps out from a small boat to the shore… As the sailor jumps forward, will the boat move? If yes, in which direction and why.

• Answer: Yes, the boat will move backwards (away from the shore/in the direction opposite to the sailor’s jump).
• Reason: According to Newton’s third law of motion, when the sailor exerts a forward muscular force on the shore to jump, the shore/ground simultaneously exerts an equal and opposite force on the sailor. Similarly, the sailor exerts a backward force on the boat, and the boat exerts an equal and opposite forward force on the sailor. Consequently, the boat experiences a net backward push and moves backward.

8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason behind it.

• Answer: The landing mat or sand bed increases the time interval over which the athlete’s momentum is brought to zero during landing.
• Reason: According to Newton’s second law, the impact force is inversely proportional to the time taken to stop (F=p/t). By lengthening the stopping time (t), the peak force (impact) experienced by the athlete’s body is significantly reduced, preventing severe injury.

9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

• Correct Option: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.
• Reason: Newton’s third law states that action and reaction forces are always equal in magnitude and opposite in direction, regardless of the masses or states of motion of the interacting objects.

10. The acceleration-mass graph for a force is plotted in Fig. 6.40. Plot the force-mass graph for this case.

• Data Extraction from Fig. 6.40:
• 
  

At m=1kg,At m = 1 kg, a = 10.0 m s^-2 implies F = m x a = 1 x 10.0 = 10 N

At m = 2 kg, a = 5.0 m s^-2 implies F = 2 x 5.0 = 10 N

At m = 4 kg, a = 2.5 m s^-2 implies F = 4 x 2.5 = 10 N

At m = 5 kg, a = 2.0 m s^-2 implies F = 5 x 2.0 = 10 N

• Conclusion for the Force-Mass Graph:
• 
  

The magnitude of the force F remains constant at 10 N across all mass values.

Plot description: The force-mass graph will be a horizontal straight line parallel to the mass axis starting at F=10N on the force axis.

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.

• Given from Fig. 6.41:
• 
  
  • Mass (m) = 10 kg
  • Initial velocity at t1=0s is u=10ms-1
  • Final velocity at t2=8s is v=30ms-1
• Calculation of Acceleration (a):a=(v-u)/(t2-t1)=(30-10)/(8-0)=20/8=2.5ms-2
• Calculation of Force (F):F=ma=10kg2.5ms-2=25N
• Answer: The force acting on the object is 25 N.

12. 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…

• Given:
• 
  
  • Mass of bullet (m) = 50 g = 0.05 kg
  • Initial velocity (u) = 100ms-1
  • Final velocity (v) = 0ms-1 (it stops)
  • Distance penetrated (s) = 50 cm = 0.5 m
• Step 1: Find acceleration using the kinematic equation (v2=u2+2as):0^2 = (100)^2 + 2 x a x 0.5
• 0 = 10,000 + a
• a = -10,000 m s^-2
• Step 2: Calculate the stopping force (F=ma):F=0.05kg(-10,000ms-2)=-500N
• Answer: The magnitude of the stopping force acting on the bullet is 500 N (acting in the direction opposite to the bullet’s motion).

13. 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…

• Given:
• 
  
  • Final/Kick velocity (v) = 108kmh-1=108(5/18)=30ms-1
  • Initial velocity (u) = 0ms-1 (assuming the ball was stationary before the penalty kick)
  • Force (F) = 800 N
  • Mass (m) = 0.4 kg
• Step 1: Find acceleration (a):F = ma => 800 = 0.4 x a => a = 800 / 0.4 = 2000 m s^-2
• Step 2: Find time of contact (t) using v=u+at:30 = 0 + 2000 x t => t = 30 / 2000 = 0.015 s
• Answer: The time of contact between the foot and the ball is 0.015 s (or 15 ms).

14. 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 is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied… distance traveled before coming to rest?

• Given:
• 
  
  • Mass (m) = 2 kg
  • Initial velocity (u) = 10ms-1
  • Final velocity (v) = 0ms-1
  • Frictional force (Ffriction) = 7 N (opposing)
  • Additional opposing force (Fadd) = 3 N (opposing)
• Step 1: Total opposing force (Fnet):Fnet=-(7N+3N)=-10N
• Step 2: Acceleration (a):a=Fnet/m=-10/2=-5ms-2
• Step 3: Distance traveled (s) using v2=u2+2as:0^2 = (10)^2 + 2 x (-5) x s
• 0 = 100 – 10s => 10s = 100 => s = 10 m
• Answer: The object travels a distance of 10 m before coming to rest.

15. A tractor pulls a harrow of mass m1 with a net force F resulting in an acceleration a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration a2. If the tractor pulls the trolley with the harrow placed on it… obtain an expression for the resulting acceleration.

• Given cases:
• 
  
  • For the harrow: F = m1 x a1 => m1 = F / a1
  • For the trolley: F = m2 x a2 => m2 = F / a2
• Combined case (trolley with harrow placed on it):
• 
  
  • Total mass (mtotal) = m1+m2
  • Force applied = F
  • Resulting acceleration (anew):anew=F/(m1+m2)
• Substitute expressions for m1 and m2:anew=F/((F/a1)+(F/a2))=F/(F(1/a1+1/a2))anew=1/((a2+a1)/(a1a2))=(a1a2)/(a1+a2)
• Answer: The expression for the resulting acceleration is anew=(a1a2)/(a1+a2).

16. When the pole of a bar magnet is brought close to a magnetic compass… both forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move. Explain why.

• Answer: According to Newton’s third law, the magnetic force exerted by the bar magnet on the compass needle is equal in magnitude and opposite to the force exerted by the needle on the bar magnet.
• Reason for motion difference: Acceleration produced depends on mass (a=F/m).
• 
  
  • The compass needle has a very small mass and is delicately pivoted on a sharp point with negligible friction, so the magnetic force produces a large enough acceleration to make it visibly rotate/move.
  • On the other hand, the bar magnet is much heavier (large mass) and is typically held by hand or resting on a surface with significant static friction/inertia, making its resulting acceleration practically zero, so it does not move.