9th Class Physics Chapter 3 (Numerical Problems with Solutions)

UNIT No. 3

Dynamics

(Numerical Problems)

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

Solution

Data

Force applied on body = F = 20 N

Acceleration produced in the body = a = 2 ms^-1

Mass of the body = m =?

Formula

sF = ma

Solution

F=ma

Rearranging this equation for m

\[m = \frac{F}{a} \] 

Putting values in this equation

\[m = \frac{20}{2} \]

 m = 10 kg     Ans 

3.2        The weight of a body is 147 N. What is its mass?

(Take the value of g as 10 ms^-2)

Solution

Data

Weight of the body = w = 147 N

Value of gravitational acceleration = g = 10 ms^-2

Mass of the body = m =?

Formula

w = mg

Solution

w=mg

Rearranging this equation for m

\[m = \frac{w}{g} \]

putting values in this equation

\[m = \frac{147}{10} \]

m = 14.7 kg  Ans 

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

Solution

Data

Mass of the body = m = 10 kg

Value of gravitational acceleration = g = 10 ms^-2

Force required to prevent the body from falling = F =?

Formula

F = ma

Solution

F=ma

Here

a = g =10 ms^-2

Putting values in the above equation

F = (10)(10)

F = 100 N             Ans

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

Solution

Data

Force applied on mass = F = 100 N

Mass of the body = m = 50 kg

Acceleration produced in the body = a = ?

Formula

F = ma

Solution

F=ma

Rearranging this equation for acceleration a

\[a = \frac{F}{m} \]  

\[m = \frac{100}{50} \]

a = 2 ms^-2      Ans 

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?

Solution

Data

Weight of the body = w = 20 N

Acceleration with which body moves upward = a = 2 ms^-2

Value of gravitational acceleration = g = 10 ms^-2

Force is required to move it vertically upward = F =?

Formula

w = mg

F = ma

Solution

Total Force required to move the body vertically upward = Reaction force + Force required produce acceleration

                            F = F_R + F_a      --------- (1)

First, we will calculate the mass of the body

w= mg

Re arranging this equation for m

\[m = \frac{w}{g} \]  

\[m = \frac{20}{10} \]

   m = 2 kg

Reaction force = F_R = w = mg = (2)(10) = 20 N

Force required produce acceleration = F_a = ma = (2)(2) = 4 N

Putting these values in equation (1)

F = F_R + F_a

     F = 20 N + 2 N

            F = 22 N       Ans   

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.

Solution

Data

Mass of 1^st body = m₁ = 52 kg

Mass of 2^nd body = m₂ = 48 kg

value of gravitational acceleration = g = 10 ms^-2

Tension in the string = T =?

Acceleration in the bodies = a =?

Formula

\[T = \frac{2m_{1}m_{2}}{m_{1}+m_{2}} \]  

\[a = \frac{m_{1} - m_{2}}{m_{1}+m_{2}} \]

Solution

First, we will calculate the tension in the string 

\[T = \frac{2m_{1}m_{2}}{m_{1}+m_{2}} \]

Putting values in this equation

\[T = \frac{2(52)(48))}{52 + 48} \] 

\[T = \frac{4992}{100} \] 

T = 499.2 N         

Rounding the answer

T = 500 N     Ans

Now, we will calculate the acceleration in the bodies

\[a = \frac{m_{1} - m_{2}}{m_{1}+m_{2}} \] 

putting values in this equation

\[a = \frac{52 - 48}{52 + 48} \] 

\[a = \frac{4}{100} \]

a = 0.04 ms^-2         Ans

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 kg mass is moving vertically downward. Find the tension in the string and the acceleration in the bodies.

Solution

Data

mass moving vertically downward = m₁ = 24 kg

mass lying over horizontal table = m₂ = 26 kg

value of gravitational acceleration = g = 10 ms^-2

Tension in the string = T =?

Acceleration in the bodies = a =?

Formula

\[T = \frac{m_{1}m_{2}}{m_{1}+m_{2}}g \]  

\[a = \frac{m_{1}}{m_{1}+m_{2}}g \]

Solution

First, we will calculate the tension in the string

\[T = \frac{m_{1}m_{2}}{m_{1}+m_{2}}g \] 

putting values in this equation

\[T = \frac{(24))(26))}{24 + 26}(10) \]

\[T = \frac{6540}{50} \]

T = 124.8 N         

Rounding the answer

T = 125 N     Ans

Now, we will calculate the acceleration in the bodies

\[a = \frac{m_{1}}{m_{1}+m_{2}}g \]

putting values in this equation

\[a = \frac{24}{24 + 26}(10)) \]

\[a = \frac{240}{50}) \]

a = 4.8 ms^-2           Ans

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

Solution

Data

Change in momentum = 𝛥P = P_f  - P_i = 20 Ns

Force = F = 20 N

Time required to change the momentum = t = ?

Formula

\[F = \frac{P_{f}- P_{i}}{t} \]

Solution

\[t = \frac{P_{f}- P_{i}}{F} \]

\[t = \frac{22}{20} \]

        t = 1.1 s         Ans 

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.

Solution

Data

Mass of block = m = 5 kg

Coefficient of friction between wood and marble = µ = 0.6

Gravitational acceleration = g = 10 ms^-1

Force of friction between wooden block and marble floor = F_s =?

Formula

F_s = µmg

Solution

F_s = µmg

Putting values in this equation

F_s = (0.6)(5)(10)

F_s = 30 N              Ans

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?

Solution

Data

Mass of body = m = 0.5 kg

Radius of Circle = r = 50 cm = 0.5 m

Speed of body = v = 3 ms^-1

Centripetal Force = F_c =?

Formula

\[F_{c} = \frac{mv^{2}}{r} \]

putting values in this equation

\[F_{c} = \frac{(0.5)(3)^{2}}{0.5} \]

\[F_{c} = \frac{(0.5)(9)}{0.5} \]

\[F_{c} = \frac{4.5}{0.5} \]

F_c = 9 N                Ans