Tag Archives: Work Energy

Work Done in Hanging a Chain

A chain with increasing mass per unit length from one end as a function of distance x from lighter end as $\lambda (x) = kx$ lying horizontally is to be hung against a vertical wall in such a manner that the external agent has to do minimum work. The min. work done by external agent is given by,

(A) $\frac {1}{6} kgl^3 $
(B) $\frac {1}{3} kgl^3 $
(C) $\frac {1}{2} kgl^3 $
(D) $\frac {2}{3} kgl^3 $

Solution

A vertically hanging chain with heavy end upwards has much more potential energy in comparison to other situations. If the heavy end is downwards, the potential energy is minimum. Minimum work would be required if increase in potential energy is minimum.

If horizontal plane is taken as the reference level for potential energy, then the initial potential energy is 0.

Let us calculate the final potential energy of vertical chain with heavy end at the bottom just touching the horizontal plane. Continue reading Work Done in Hanging a Chain

A particle is released from height H above the ground ….

A particle is released from height H above the ground. At a certain instant its kinetic energy is three times its potential energy. The height h above the ground and the speed of the particle v at that instant are respectively:

(1) $\frac {H}{4},\frac {3gH}{2}$
(2) $\frac {H}{4},\frac {\sqrt {3gH}}{2}$
(3) $\frac {H}{2},\frac {\sqrt {3gH}}{2}$
(4) $\frac {H}{4},\sqrt {\frac {3gH}{2}}$ Continue reading A particle is released from height H above the ground ….

An engine is attached to a wagon through a shock absorber ….

An engine is attached to a wagon through a shock absorber of length 1.5 m. The system with a total mass of 40,000 kg is moving with a speed of $72 kmh^{-1}$ when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, the spring constant is _ _ _ _ $\times 10^5 Nm^{-1}$.

Solution

10 % of K.E. = $\frac {1}{2} kx^2 $

$\therefore 0.1 \times \frac {1}{2} mv^2 = \frac {1}{2} kx^2 $

$\therefore 0.1 \times 40,000 \times (72 \times \frac {5}{18})^2 = k \times 1.0^2 $

$\Rightarrow k = 4,000 \times 20^2 = 16 \times 10^5 Nm^{-1} $

Answer: 16