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
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
For a body of mass 1 kg U-x graph is shown in If the body is released from rest at x = 2m, then its speed when it crosses x = 5m is
From Principal of Conservation of Energy
Ui + Ki = Uf + Kf
10 +0 = 2 + ½v²
⇒ v ² = 16 ⇒ v = 4 m/s
For a spring force-compression graph is shown in figure. A body of mass 5 kg moving with a speed of 8 ms–1 collides the The maximum compression in the spring is
Using this graph we can find value of Spring constant.
F = - K.x
So, K = 80 N/m
Kinetic Energy of the block will convert into spring potential energy
so, ½mv² = ½ k x²
solving x = 2 m
A body of mass 2 kg collides with a massless spring of force constant K = 4 N/m. The spring compresses by 1m. If coefficient of friction between the body and the surface is 0.1, the speed of the body at the time of collision is :
The Kinetic Energy of the object will go to increase potential energy of the spring and a part of energy will be lost to friction
U1= ½ K x²= ½ × 4× 1² = 2J
Work done against friction = ü.mg.x= 0.1× 20 × 1= 2 J
So total kinetic energy is 4 = ½× 2 ×v ² ⇒ v = 2 m/s
A system absorbs 600 J of the system works equivalent to –900 J by The value of ΔE for the system is :
Work done by external agent = -900 J and energy supplied is 600 J So, change in kinetic energy is -300 J
A particle is released from height \(S\) from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of Earth and the speed of the particle at that instant are respectively:
(2021)
Let the initial height be \(S\). At height \(h\), \(KE = 3PE = 3mgh\). By conservation of energy, \(mgS = mgh + 3mgh = 4mgh\), so \(h = S/4\). Also, \(KE = \frac{1}{2}mv^2 = 3mgh\), substituting \(h\) gives \(v^2 = 6g(S/4) = \frac{3gS}{2}\), so \(v = \sqrt{\frac{3gS}{2}}\).
A child is sitting on a swing. Its minimum and maximum heights from the ground is \(0.75\text{ m}\) and \(2\text{ m}\) respectively, its maximum speed will be:
(2001)
Maximum speed occurs at minimum height, where potential energy is lowest and kinetic energy is highest. Minimum speed (zero) occurs at maximum height. By conservation of mechanical energy: \(mgh_{max} + \frac{1}{2}mv_{min}^2 = mgh_{min} + \frac{1}{2}mv_{max}^2\). With \(v_{min}=0\), \(mg(2) = mg(0.75) + \frac{1}{2}mv_{max}^2\). \(2g - 0.75g = \frac{1}{2}v_{max}^2 \Rightarrow 1.25g = \frac{1}{2}v_{max}^2\). Using \(g = 10\text{ m/s}^2\), \(v_{max}^2 = 2.5 \times 10 = 25\), so \(v_{max} = 5\text{ m/s}\).