Solutions to Exercises on Newton's Laws of Motion HC Verma's Concepts of Physics Part 1 (1-6)

 Q#1

A block of mass 2 kg placed on a long friction-less horizontal table is pulled horizontally by a constant force F. It is found to move 10 m in the first two seconds. Find the magnitude of F.

Answer:  
First of all we need to find the acceleration of the block. Here time t = 2 s, distance s = 10 m, initial velocity u = 0 m/s, acceleration a =? using the formula 

s = ut + ½at², we get   
10 = 0 + ½a(4)
2a = 10 
a = 5 m/s² 

Given that mass m = 2 kg, From the relation F = ma, we get,
F = (2 kg)(5 m/s²) = 10 N

Q#2      
A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it?   

Answer:
To know the force we need to know both mass and acceleration. Mass m = 2000 kg (given) but acceleration (-ve here) has to be calculated from the given data from the equation of kinematics

v² = u² + 2as. 

v = 0 m/s, s = 4 m, u = 40 km/h = 40000m/3600s = 100/9  m/s,

So, 0 = (100/9)² + 2a(4)

a = -10000/(81 x 8) = -1250/81 m/s²

Now force F = ma = (2000)(-1250/81) N = -2500000/81 N = -3.1 x 10⁴ N

(Negative signs of force and acceleration show that their direction is opposite to the direction of initial velocity.)         

Q#3
In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of 5 x 10⁶m/s in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is 9.1 x 10^-31 kg.

Answer:
Initial velocity u = 0 m/s, Final velocity v = 5 x 10⁶ m/s, Distance traveled s = 1 cm = 0.01 m, acceleration a =?

using v² = u² + 2as

(5 x 10⁶)² = 0² + 2a(0.01)

0.02a = 25 x 10¹²

a = 12.5 x 10¹⁴ m/s² 

Given that mass m = 9.1 x 10^-31 kg

Now Force
F = ma = (9.1 x 10^-31 kg)(12.5 x 10¹⁴ m/s²)
F = 113.75 x 10^-17 N ≈ 1.1 x 10^-15 N     

Q#4        
A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of 0.3 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s².        

Answer:
Let us consider the lower block. Force due to gravity which is weight = mg = 0.3 x 10 N = 3 N. To balance it tension in the string = 3 N.

Now consider the upper block and string. Weight of upper block = 0.2 x 10 N = 2 N. Pull by lower string = 3 N, Both downwards.

So, to balance it the string will pull them upwards with a force = 2 N + 3 N = 5 N. Hence tension in upper string = 5 N. The picture below will explain it further.

Q#5            
Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.  

Answer:
Due to force F both the blocks will be pulled. Acceleration of the blocks = F/2m. The inertia of block at farther end will resist this acceleration with a force = mass of that block x acceleration = m(F/2m) = F/2. So the tension in the string = F/2.    

Q#6
A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.

Answer:
Mass of the particle = 50 g = 0.05 kg. 
To know the force acting on the particle, we need to know the acceleration of the particle. In the velocity-time graph the slope of the curve gives the instantaneous acceleration. Let us find it for the given instant.   

At t = 2 seconds

The graph is a straight line with positive slope. It means the particle has a constant acceleration with magnitude = 15/3 = 5 m/s².

So the force acting on it = mass x acceleration = 0.05 kg x 5 m/s² 
= 0.25 N, it acts along the motion because it is positive.

At t = 4 seconds   

The graph is horizontal to time axis meaning thereby the velocity is constant and no acceleration. It can be understood in this way too that the slope which represents acceleration is zero. Since there is no acceleration at t = 4 s. so there is no force acting on the particle at this instant, Force= zero. 
   
At  t = 6 seconds

The graph shows that velocity is uniformly decreasing with the time and the acceleration which is represented by the slope is negative. From the graph value of acceleration = -15/3 = -5 m/s².
Force = mass x acceleration = 0.05 x (-5 N) = -0.25 N

So, the force acting on the particle is 0.25 N and negative sign shows that its direction is opposite to the motion.

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