Find a complete solution for chapter 3 “Dynamics” numerical problems. Fine grip over physics concepts along with formula are necessary to solve problems in physics.

For more visit

**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}?

^{– 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: **m_{1} = 52 kg and m_{2} = 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: : **m_{1} = 24 kg and m_{2} = 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}?

^{– 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*