Loading....
Coupon Accepted Successfully!

 

Illustrations

Example-1

A vehicle is travelling at a velocity of 10 m s–1. An accelerating force of 240 N is applied for a duration of 15 s. The mass of the vehicle is 1200 kg. What is the final velocity of the vehicle?

Solution

u = 10 m s–1F = 240 N, t = 15 s, m = 1200 kg, a = ?, v = ?

 

F = ma or Description: 51515.png

 

v = u + at

 

= 10 m s–1 + (0.2 m s–1  15 s)

 

= 10 + 3 = 13 m s–1
 

 

Example-2

A gun having a mass of 5000 g fires a 20 g bullet at a speed of 500 m s–1. Find the recoil velocity of the gun.

Solution

Mass of the bullet, m1 = 20 g = 0.02 kg

 

Velocity of the bullet, v1 = 500 m s–1

 

Mass of the gun, m2 = 5000 g = 5 kg

 

Let the recoil velocity of gun be v2 m s–1.

 

According to the law of conservation of momentum,

 

Momentum of bullet = Momentum of gun

 

m1 v1 = m2 v2

 

0.02  500 = 5  v2

 

Description: 51527.png
 

 

Example-3

A book of mass 0.5 kg lies on the surface of a table. The coefficient of friction between the book and the surface of the table is 0.6. Calculate the force required to just move it (take g = 10 m s–2).

 

Another identical book is kept on the book on the table. How is the force needed to move it affected?

Solution

Given m = 0.5 kg,

 

R = W = mg = 0.5 × 10 = 5N, μ = 0.6

 

In the limiting condition,

 

Force required = Force of friction

 

​∴ Force required F = f = μR = 0.6 5=3Ν

 

When another book is placed over the first book, the reaction becomes

 

R´ = W´ = 2W

 

= 2 0.5 10 = 10 N

 

Force required F´ = f´ = μR´

 

= 0.6 10=6Ν
 

 

Example-4

A block of weight 10 N lies on a rough table. It just moves when a force of 7 N is applied on it. Calculate the coefficient of limiting friction between the block and the table.

Solution

In the limiting condition, force of friction f = applied force F = 7 N,

 

Normal reaction, R = weight W = 10 N

 

From the relation f = μR,

 

Coefficient of friction Description: 51552.png
 

 
Example-5

Two bodies each of mass 25 kg are separated by a distance 1 m. Find the force between them.

 

What will be the force, if the distance between the masses is doubled?

 

[Assume G = 6.673 × 10−11 N m2 kg–2].

Solution

Mass of the first body = m1 = 25 kg

 

Mass of the second body = m2 = 25 kg

 

Separation between them = r1 = 1 m

 

We know that, the gravitational force

 

Description: 51570.png

 

F1 = 4.17 × 10−8 Ν

 

For the second part r2 = 2r1

 

Description: 51595.png

 

F2 = 1.04 × 10−8 Ν
 


Example-6

Calculate the values of the acceleration due to gravity (i) at a depth of 8 km and (ii) at an altitude of 32 km from the surface, given that the value of g on the surface at sea level is 9.8 m s–2.

Solution

(i) The value of g at any depth is given by

Description: 51623.png

 

Description: 51631.png

 

Description: 51640.png

 

= 9.788 m s–2


(ii) The value of g at an altitude of 32 km is

 

Description: 51651.png

 

Description: 51658.png

 

Description: 51666.png

 

g = 9.70 m s–2
 





Test Your Skills Now!
Take a Quiz now
Reviewer Name