Chapter# 3
Dynamics
Numerical Question Solution (slo)
National Book Foundation
List of formulas and units related to chapter#3 that
you need to memorize before moving towards solution portion . You will get the
basic idea of using different formulas for same quantity .
Formulas
·
F=ma
·
F=mg
·
Fc=(mv^2)/r
·
P=mv
·
F=P/t
·
a=(m1-m2)g/(m1+m2)
·
T=2(m1m2)g/(m1+m2)
Units
F -> N
P -> Ns
T - > N
Numerical Solution
1. A vehicle is accelerating at 5 m/s^2under the action of 2500N force, What is mass of the vehicle ?
Given
a= 5 m/s^2
F=2500N
To Find
m=?
Calculation
From second equation of motion
F=ma
2500=m(5)
m=2500/5
m=500 kg
2. A boy is holding a book of mass 2kg . How much force is he applying on the book ? If he moves it up with acceleration of 3 m/s^2 , How much total force should he apply on the book ?
Given
m=2kg
a= 3m/s^2
To find
Total force = ?
Calculation
Force required for holding book
F1=mg=(2)(9.8)= 19.6N
From Second equation of motion
F2=ma =2*3=6N
Total required force =F1+F2
= 19.6+6
=25.6 N
3. A girl of mass 30 kg is running with velocity of 4 m/s . Find her momentum .
Given
m=30 kg
v=40 m/s
To find
P=?
Calculation
P=mv
=30*40
=1200Ns
4.A 2 kg steel ball is moving with speed of 15 m/s. It hits with bulk of sand and comes to rest in 0.2 seconds . Find force applied by sand bulk on the ball.
Given
m=2kg
v=15 m/s
t=0.2s
Total Find
F = ?
Calculation
a=v/t
=15/0.2 =75 m/s^2
F=ma
=2(75)
=150 N
5. A soccer player kicks the football with force of 1200 N with impact time of 0.60 seconds . Find the change in momentum of football ?
Given
F=1200 N
t=0.60
To find
P=?
Calculation
F=P/t
P=F*t
=1200 * 0.60
P=720 N
6. A 100 grams bullet is fired from 5 kg gun . Muzzle velocity of bullet is 20 m/s . Find recoil velocity of the gun .
Given
mass of bullet (m) =100=(100/102//00 )kg=1/10=0.1kg
velocity of bullet (v)=20m/s
mass of gun (M)=5
To find :
velocity of recoil of gun (V)=?
Calculation
as gun and bullet is at rest initially so initial momentum is zero
Final momentum = momentum of bullet + momentum of gun
=mv+MV
=0.1(20)+5V
=2+5V
Applying law of conversation of momentum
initial momentum = final momentum
0=2+5V
5V=-2
V=-2/5 = -0.4 m/s
Negative sign shows that the direction of recoil velocity of the gun is opposite to the direction of the velocity of the gun .
7. A robotic car of 15kg is moving with 25 m/s . Brakes are applied to stop it . Brakes apply constant force of 50 N . How long does the car take to stop .
Given
mass (m) =15 kg
velocity (v) = 25 m/s
Force (F) = 50 N
To find
t=?
Calculation
P=mv = 15 * 25
=375 Ns
F=P/t
t=P/F
=375/50
=7.5s
8.A force of 20 N is applied on stack of books of mass 3 kg to push it on a rough table surface with uniform speed of 20 m/s . How much net force is acting on it ? What is the value of friction between the stack of books and the table surface ?
Given
F= 20 N
m=3kg
v= 20 m/s
To find
net force ?
coefficient of friction =?
Calculation
9. A wooden block of mass 50 kg is placed on the rough surface of table . What is the weight of the block . What is normal force on the block by table ? If coefficient of static friction between the block and the table is 0.4 , find the minimum force to move the block on the table .
Given
m=50 kg
coefficient of static friction = 0.4
To find
w=?
N=?
F=?
Calculation
w=mg = 50 *9.8
= 490 N
As normal force = w = 490 N
coefficient of static friction = f/N
0.4= f/ 490
f=0.4*490 =196 N
10. A cyclist weighing 500 N is moving in circular track of radius 60 meters with a uniform speed of 12 m/s . Find the centripetal force required to keep him moving in this circular track .
Given
w= 500 N
r= 60 m
v= 12 m/s
To find
Fc=?
Calculation
w=mg
500=m(9.8)
500/9.8=m
m=51.02
Fc=(mv^2)/r
=(51*12*12)/600
=7344/60
=122.4 N
11. A stone of 500 g tied to a string of length 50 cm is whirled by a girl in a horizontal circle with speed of 3 m/s . Find the tension in the string. Also tell which force is responsible to produce tension in the string .
Given
mass =500g=(500/1000 )kg =0.5 kg
radius =50cm=50/100 =0.5 m
v= 3m/s
To find
Fc= ?
Calculation
Fc=(mv^2)/r = (0.5*3*3)/0.5 =9 N
=
12. Two bodies of masses 10 kg and 8 kg are hanging by string that is passing over friction less pulley . Both bodies move vertically on releasing them . Find the acceleration of the system and tension produced in the string ?
Given
m1= 10kg
m2= 8kg
To find
a=?
T=?
Calculation
a=(m1-m2)*g/(m1+m2)
=(10-8)9.8/(10+8)
=(2*9.8)/18
=19.6/18 =1.1 m/s^2
T=2(m1m2)*g/(m1+m2)
=2(10*8)*9.8/(10+8)
=1568/18
=87.1 N
13. Find acceleraion due to gravity using atwood machine having masses 5kg and 3 kg if system has acceleration of 2m/s^2 .
Given
m1=5kg
m2=3kg
a=2m/s^2
To find
g=?
Calculation
a=(m1-m2)*g/(m1+m2)
2=(5-3)*g/(5+3)
2=2g/8
2*8=2g
16=2g
g=16/2=8 m/s^2
FAQ
Q.1 What is value of g ?
Ans . 9.8 m/s^2
Q.2 Is weight is force ?
Ans . Yes weight is consider normal force
Q.3 What is tension and its unit ?
Ans . Tension is type of force. So it's SI unit is N .
Q.4 What is symbol of Tension ?
Ans T
Q.5 Which type of force is Tension ?
Ans It's a pulling force .