A child on a swing is 1 m above the ground at the lowest point and 6 m above the ground at the highest point. The horizontal speed of the child at the lowest point of the swing is approximately
From Principal of conservation of Energy in absence of non-conservative forces,
Ui+ Ki= Uf+ Kf
⇒mg(6) +0 = mg(1) + ½mv²
⇒mg(5)= ½mv²
⇒ v²=100 or v =10 m/s
A mass m slips along the wall of a hemispherical surface of radius R. The velocity at the bottom of the surface is
From Principal of conservation of Energy in absence of non-conservative forces,
Ui+ Ki= Uf+ Kf
mgR+0= 0+ ½mv²
v= √2gR
The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is
From Principal of conservation of Energy in absence of non-conservative forces,
Ui+ Ki= Uf+ Kf
½KL²+ 0 = ½mv²
⇒v= √(KL²/m)
Momentum is P= m.v = m. √(KL²/m) =(mK)½ L
A simple pendulum is released from A as shown. If m and l represent the mass of the bob and length of the pendulum, the gain in kinetic energy at B is
From Principal of conservation of Energy in absence of non-conservative forces,
Ui+ Ki= Uf+ Kf
⇒ Ui- Uf= Kf - Ki
⇒ Loss in Potential Energy = Gain in Kinetic Energy
⇒ Gain in Kinetic Energy = m.g. l .cos 30° = √3mgl/2
The position of a particle of mass 1 kg moving along x-axis at time t is given by x=t²/2. The work done by the force from t=0 to t=3 sec
Given, x = t²/2 differentiating v=dx/dt= t
Initial Speed Vi= 0 m/s
Final Speed Vf= 3 m/s
From Work Energy Theorem, Work done is equal to change in kinetic energy
So, Work Done = ½m (Vf² - Vi²)= 1/2 × 1× (9-0) = 4.5 J
A force act on a 2 kg particle such a way that position of the particle as a function of time is given by x = 3t -4t²+t³, Where x is in meter and t is in second. Work done during first 4 second is
Given x = 3t -4t²+t³ differentiating v = dx/dt = 3-8t+ 3t²,
Initial speed = 3 m/s
Final speed = 3-32+48= 19 m/s
From Work Energy Theorem,
Work Done = Change in Kinetic Energy= ½×2×(19²-3²)= 352 J
A 5 kg ball when falls through a height of 20 m acquires a speed of 10 m/s. Find the work done by air resistance
From work energy theorem, work done by all the forces is equal to change in kinetic energy.
So, Work done by gravity + Work done by air resistance = Final Kinetic Energy - Initial Kinetic Energy
5×10×20 + Work done by air resistance = 1/2× 5 ×(10²-0)
⇒ 1000 + Work done by air resistance = 250
⇒ Work done by air resistance = 250 -1000= -750 J
The potential energy of a particle oscillating on x-axis is given as : U = 20 + (x – 2)2 Joules. If total mechanical energy of the particle is 36 J, then its maximum kinetic energy is:
Given U = 20 + (x – 2)2
For Equilibrium , dU/dx = 2 (x-2) =0, So, x=2 is the point of Equilibrium.
d²U/dx²= 2 >0 so, x= 2 is the point of stable equilibrium.
U min = 20 J
As Total Energy is 36 J.
U min + K max = 36 J
⇒ K max = 16 J
A person draws water from a 5m deep well in a bucket of mass 2 kg of capacity 8 litre by a rope of mass 1 kg. What is the total work done by the person ? [g = 10 m/s2]
Work Done in Pulling the Bucket + Rope System=
Mgh+mg(h/2)
10×10× 5 + 1×10×(5/2) =525 J
A bullet of mass 10g is fired horizontally with a velocity 1000 ms–1 from a rifle situated at a height 50m above the If the bullet reaches the ground with a velocity 500 ms–1, the work done against air resistance on the bullet is : (g = 10 ms–2)
From Work Energy Theorem,
Total Work Done = Change in Kinetic Energy
Work Done by Gravity + Work Done by Air Resistance = ½m (Vf 2 - Vi 2 )
0.001 × 10×50 + Work done by Air Resistance = ½× 0.001×( 500² - 1000²)
Solving We get, Work Done by Air Resistance = - 3755 J