Numerical Problems

 

 

3.1 A force of 20 N moves a body with an acceleration of 2 ms ^{– 2} . What is its mass?

(10kg)

Solution:   Force = F = 20 N

                     Acceleration = a = 2 ms ^{– 2}

                                 Mass = m = ?

F = ma

Or               m =

m =    = 10 kg

 

 

 

3.2 The weight of a body is 147 N. What is its mass? (Take the value of g as 10 ms ^{– 2} )

 (14.7 kg)

Solution:      Weight = w = 147 N

                     Acceleration due to gravity = g = 10 ms ^{– 2}  

                     Mass = m = ?

w = mg

or                         m =

m =

m = 14. 7 kg

 

 

 

 

 

3.3 How much force is needed to prevent a body of mass 10 kg from falling?

 (100 N)

Solution:     Mass = m = 50 kg

                    Acceleration = a = g = 10 ms ^{– 2}  

                    Force = F = ?

F = m a

F = 10 x 10

F = 100 N

 

 

 

3.4 Find the acceleration produced by a force of 100 N in a mass of 50 kg.

 (2 ms ^{– 2} )

Solution:     Force = F = 100 N

                   Mass = m = 50 kg

Acceleration = a = ?

F = m a

Or             a =

a =

a = 2 ms ^{– 2} 

 

 

 

3.5 A body has weight 20 N. How much force is required to move it vertically upward with an acceleration of 2 ms ^{– 2}?

(24 N)

Solution:          Weight = w = 20 N

                        Acceleration = a = 2 ms ^{– 2}

Vertically upward force (Tension) = T = ?

F_net  = T – w

Or                     ma = T – mg

Or                     ma + mg = T

Or                     T = m (a + g) ……………………(i)

Now,                 m =

m =   = 2 kg

Putting the value of m in Eq.(i), we get

T = 2 (2 + 10)

= 2(12)

T = 24 N

 

 

 

 

 

3.6 Two masses 52 kg and 48 kg are attached to the ends of a string that passes over a frictionless pulley. Find the tension in the string and acceleration in the bodies when both the masses are moving vertically.

(500 N, 0.4 ms ^{– 2}  )

Solution:           m₁ = 52 kg       and       m₂ = 48 kg

• Tension                    T = ?
• Acceleration         a = ?

 

• T =    g

             T =    × 10

            T =

T = 499.20 ≈ 500 N

• a = g

 

 a =   × 10

 a =   × 10

                 a  = 0.4 ms ^{– 2}

 

 

 

3.7 Two masses 26 kg and 24 kg are attached to the ends of a string which passes over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 N mass is moving vertically downward. Find the tension in the string and the acceleration in the bodies.  

(125 N, 4.8 ms ^{– 2})

Solution: :           m₁ = 24 kg       and       m₂ = 26 kg

• Tension                    T = ?
• Acceleration a = ?

 

• T =    g

             T =    × 10

            T =

T = 124.8 ≈ 125 N

• a = g

 

 a =   × 10

 a =   × 10

                 a  = 4.8 ms ^{– 2}

 

 

 

3.8 How much time is required to change 22 Ns momentum by a force of 20 N?

 (1.1s)

Solution:       Change in momentum = P_f – P_I = 22 Ns

                      Force = F = 20 N

Time = t = ?

F =

t =

t =  = 1.1 S

 

 

 

 

 

3.9 How much is the force of friction between a wooden block of mass 5 kg and the horizontal marble floor? The coefficient of friction between wood and the marble is 0.6.

 (30 N)

Solution:         Mass = m = 5 kg

                       Coefficient of friction = µ = 0.6

Force of friction = F_S = ?

F_S = µ R                               (where R = mg)

F_S = µ mg

F_S = 0.6 x 5 x 10 = 30 N

 

 

 

3.10 How much centripetal force is needed to make a body of mass 0.5 kg to move in a circle of radius 50 cm with a speed 3 ms ^{– 1}?

 (9 N)   

Solution:          Mass = m = 0.5 kg

                        Radius of the circle = r = 50 cm =  = 0.5 m

Speed = v = 3 ms ^{– 1}

Centripetal force = F_c = ?

                                   F_c =

                                  F_c =

                                 F_c =    = 9 N