Dynamics - Numerical… | Class 09 Physics

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               $m = \frac{F}{a}$                   $m =  \frac{20}{2}$  = 10 kg

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 = mgor                         m =  $\frac{w}{g}$                             m =  $\frac{147}{10}$                              m = 14. 7 kg

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 = $10 \times10$                     F = 100 N

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 aOr             a =  $\frac{F}{m}$                a =  $\frac{100}{50}$                 a = 2 $ms^{-2}$

Does a body 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 = ?                        Fnet  = T – w Or                     ma = T – mgOr                     ma + mg = TOr                     T = m (a + g) ……………………(i)Now,                  m = $\frac{w}{g}$                           m =$\frac{20}{10}$= 2 kgPutting the value of m in Eq.(i), we get                         T = 2 (2 + 10)                            = 2(12)                         T = 24 N

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 kgTension                T =?   Acceleration a =?      T =$\frac{ (2m_1 m_2)}{(m_1+ m_2 )}$g             T = $\frac{(2 ×52 ×48)}{(52 + 48)}\times$ 10            T =  $\frac{49920}{100}$             T = 499.20 ≈ 500 Na =$\frac{(m_1  – m_2)}{(m_1+ m_2 )}g$                  a = $\frac{(52 - 48)}{(52 + 48)}\times 10$               a =$\frac{4}{100}\times 10$                 a  = 0.4 $ms^{-2}$

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 =$\frac{(m_1 m_2)}{(m_1+ m_2 )}g$             T = $\frac{(24 ×26)}{(24 + 26)}\times$ 10            T = $\frac{6240}{50}$             T = 124.8 ≈ 125 N a =$\frac{(m_1)}{(m_1+ m_2 )}$ g                a = $\frac{24}{(24 + 26)}\times10$               a = $\frac{24}{50}\times10$               a  = 4.8 $ms^{-2}$

How much time is required to change 22 Ns momentum by a force of 20 N? (1.1s)

Solution:       Change in momentum = Pf – PI = 22 NsForce = F = 20 NTime = t = ?F =$\frac{(P_f  – P_i)}{t}$  t =$\frac{(P_f  – P_i)}{F}$ t =$\frac{22}{20}$= 1.1 S

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 marble is 0.6. (30 N)

Solution:         Mass = m = 5 kg                       Coefficient of friction = µ = 0.6                        Force of friction = FS = ?                        FS = µ R                      (where R = mg)                        FS = µ mg                                            FS = $0.6 \times 5 \times 10$ = 30 N

How much centripetal force is needed to make a body of mass 0.5 kg move in a circle of radius of 50 cm with a speed of 3 $ms^{-1}$? (9 N)

Solution:          Mass = m = 0.5 kg                        Radius of the circle = r = 50 cm =$\frac{50}{100}$= 0.5 m                        Speed = v = 3 $ms^{-1}$                        Centripetal force = Fc = ?                                   Fc =$\frac{(mv)^{2}}{r}$                                   Fc =$\frac{(0.5 \times3)^{2}}{0.5}$                                  Fc = $\frac{(0.5 \times9)}{0.5}$=$\frac{4.5}{0.5}$= 9 N

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