Question

A lift of mass $M =500 kg$ is descending with speed of $2 ms ^{-1}$ Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 ms ^{-2}$ The kinetic energy of the lift at the end of fall through to a distance of $6 m$ will be ___ $kJ$

Hint:

The kinetic energy of an object is calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the final velocity.

Answer 1

Correct Answer: 7

The correct answer is 7.
$v^2=u^2+2as$
$=2^2+2(2)(6)$
$=4+24=28$
$KE=\frac{1}{2}m{v}^2$
$=\frac{1}{2}(500)28$
$=7000J$
$=7kJ$

Answer 2

The velocity of the lift at the end of the fall is found using the equation of motion:

\[ v^2 = u^2 + 2as \]

where:

• \( u = 2 \, \text{m/s} \),
• \( a = 2 \, \text{m/s}^2 \),
• \( s = 6 \, \text{m} \).

Substituting the values:

\[ v^2 = 2^2 + 2 \times 2 \times 6 = 4 + 24 = 28 \]

\[ v = \sqrt{28} = 5.29 \, \text{m/s} \]

The kinetic energy \( KE \) is given by:

\[ KE = \frac{1}{2} m v^2 \]

Substituting the values:

\[ KE = \frac{1}{2} \times 500 \times 28 = 7000 \, \text{J} = 7 \, \text{kJ} \]

Thus, the kinetic energy of the lift is 7 kJ.