Question 1: A 5kg monkey swings from a branch to another branch of 2m higher. What is the change in potential energy? What is the initial velocity of initial kinetic energy of the monkey?
Solution:
Here, we have,
Mass of monkey (m)=5kg
Increase height (h)=2m
acceleration due to gravity (g)=10m/sec
change in potential energy (P.E)=?
Initial velocity (u)=?
Initial kinetic energy (k.e)=?
Now,
Change in potential energy (p.e)=(mgh)upper-(mgh)lower
=5kg*10m/sec^2*2m-0
=100 joule
Again,
Distance covered (s)=2m
Final velocity (v)=0 m/sec
acceleration due to gravity (g)=-10m/sec^2
Initial velocity (u)=?
By using formula,
v^2=u^2+2as
Or, 0^2=u^2-2*10*2
Or, u^2=40
i.e. u=6.32m/sec^2
Initial velocity of initial kinetic energy of the monkey is 6.32m/sec^2.
Hence, Kinetic energy of monkey (k.e)=mv^2/2
=5kg * (6.32)^2/c
=100 joule
Question 2: A stationary mass explode into two mass 4unit and 40unit respectively. If the larger mass has an initial kinetic energy of 100joule. What is the initial kinetic energy of small mass?
Solution:
Here, we have,
Mass of smaller stationary (m1)=4unit
Mass of larger stationary (m2)=40unit
Kinetic energy of larger mass (e2)=100J
Kinetic energy of smaller mass (e1)=?
By using formula,
e1/e2=m1/m2
Or, e1=m2*e2/m1
Or, e1=40unit * 100J/4unit
Or, e1= 1000J
Hence, Kinetic energy of smaller mass (e1)=1000J
Question 3:A pump set is used to lift water to a reservoir of 6000liters capacity over an average height of 10m. If it takes 1hour to fill the reservoir completely. Calculate the work done and power supplied.
Solution:
Volume of water reservoir (v)=6000liters=6m^3
Mass of water in 1hrs (m)=v * density = 6m^3 * 1000kg/m^3=6000kg
Height of reservoir (h)=10m
time taken (t)=1hrs=3600sec
Work done in 1hrs=f*d= -mg*h
=-6000*10*10
=-6*10^5J
W=6*10^5 (Only magnitude)
Finally,
Power (p) =W/t
=6*10^5/3600
=250watt
Question 4: A 100kg car has a maximum power out of 100hp. How steep a hill can it climb up at a constant speed of 10km/hrs. If the friction force is 500N?
Solution:
Here, we have,
Mass of car (m)=100kg
Power of car (p)=100hp=74600watt
Height of hill (h)=?
Angle of inclination (Ɵ)=?
Now,
Total force applied (F)=mgSinƟ + Fc
Or, P/v=100kg*10SinƟ +500N
Or, 74600/60=1000SinƟ + 500N
Or, 74600/(50/3)=500(2SinƟ+1)
Or, 74600 * 3/50*500= 2SinƟ + 1
Or, 2SinƟ= (2238/250)-1
Or, SinƟ=1988/500
SinƟ=3976/10000
Height of hill (h)= 3976m
and Ɵ=23.43
Question 5: A railway truck of mass 4 *10^4kg moving with a velocity of 3m/sec collides with another truck of mass 2*10^4 which is at rest. The coupling joint and they moves together. what fraction of first truck initial kinetic energy to kinetic energy of both trucks after collision?
Solution:
Here, we have,
Mass of railway truck (m1)=4*10^4
Velocity of railway truck (u1)=3m/sec
Mass of another truck (m2)=2*10^4
Velocity of another truck (u2)=0m/sec (at rest)
Since, they collide to each other during collision. so final velocity are equal (v1=v2=v)
Now,
m1u1+m2u2=m1v+m2v
Or, v=(m1u1+m2u2)/(m1+m2)
Or, v= 12*10^4/6*10^4
i.e. v= 2m/sec
Initial kinetic energy of first railway truck (E1)=m1u1^2/2
=4*10^4*3^2/2
=18*10^4Joule
Kinetic energy of both truck (E2)= (m1+m2)v2^2
=(4*10^4+2*10^4)*2^2
=12*10^4Joule
Again,
E2/E1=12*10^4/18*10^4
=2/3
Question 6: A bullet of mass 20gram traveling horizontally at 100m/sec embedded itself in the center of a block of wood of mass 1kg which is suspended by a light vertical string 1m length. Calculate maximum inclination of the string to the vertical. (assuming g=9.8m/sec^2)
Solution:
Here,we have,
Mass of bullet (m1)=20gram=0.02kg
Velocity of bullet (u1)100m/sec
Mass of wood (m2)=1kg
Velocity of wood (u2)=0m/sec (at rest)
Since, bullet is embedded in wood and moves with same velocity. so final velocity is v (i.e. v1=v2=v)
m1u1+m2u2=m1v+m2v
Or, v=(m1u1+m2u2)/(m1+m2)
Or, v=1.9m/sec
Again,
(v2)^2=(u2)^2+2as
Or, (1.96)^2=0^2+2*9.8*S
Or, S=0.196m
Then,
CosƟ=OA//OB^1
Or, OB^1CosƟ=OA
Or, OB^1CosƟ=OA
Or, l*CosƟ =(OB-AB)
Or, 1*CosƟ=(1-0.196)
Or, CosƟ=0.804
i.e. Ɵ=36.48
Thus, Required maximum inclination of the string to the vertical=36.48
Thanks
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