Table of Contents
A force of 20 N moves a body with an acceleration of 2 $ms^{-2}$. What is its mass? (10kg)
Difficulty: Easy
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)
Difficulty: Easy
Solution:
Weight = w = 147 N
Acceleration due to gravity = g = 10 $ms ^{-2}$
Mass = m = ?
w = mg
or 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)
Difficulty: Easy
Solution:
Mass = m = 50 kg
Acceleration = a = g = 10 $ms ^{-2}$
Force = F = ?
F = m a
F = $10 \times10$
F = 100 N
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Find the acceleration produced by a force of 100 N in a mass of 50 kg. ($2 ms ^{-2}$)
Difficulty: Easy
Solution:
Force = F = 100 N
Mass = m = 50 kg
Acceleration = a = ?
F = m a
Or 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)
Difficulty: Easy
Solution:
Weight = w = 20 N
Acceleration = a = 2 $ms ^{-2}$
Vertically upward force (Tension) = T = ?
Fnet = T – w
Or ma = T – mg
Or ma + mg = T
Or T = m (a + g) ……………………(i)
Now, m = $\frac{w}{g}$
m =$\frac{20}{10}$= 2 kg
Putting 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. $(500 N, 0.4 ms^{-2})$
Difficulty: Easy
Solution:
m1 = 52 kg and m2 = 48 kg
Tension 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 N
a =$\frac{(m_1 – m_2)}{(m_1+ m_2 )}g$
a = $\frac{(52 - 48)}{(52 + 48)}\times 10$
a =$\frac{4}{100}\times 10$
a = 0.4 $ms^{-2}$
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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 $ms^{-2}$)
Difficulty: Medium
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)
Difficulty: Easy
Solution:
Change in momentum = Pf – PI = 22 Ns
Force = F = 20 N
Time = 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)
Difficulty: Easy
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
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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)
Difficulty: Easy
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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