Question

(a) In 1975 the roof of Montreal’s Velodrome, with a weight of 360 kN, was lifted by 10 cm so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised 4000 N (about of the car’s weight) by 5.0 cm, how much work did her force do on the car?

Short answer

1. The work done on the roof by the forces making the lift is 36 kJ.
2. The work done on the car is $2\times {10}^2\mathrm{J}$.

Step-by-step solution

Given data:

Weight of lift, F₁=360 kN

The weight lifted by, d₁=10 cm=0.1m

The weight of the car, F₂=4000 N

The car was lifted by, d₂=5.0 cm=0.05 m.

Understanding the concept:

The work-energy theorem states that the net work done by forces on an object is equal to the change in its kinetic energy.

Using the work-kinetic energy theorem, you have

$\begin{array}{rcl}\mathbf{∆}\mathbf{K}& \mathbf{=}& \mathbf{W}\\ & \mathbf{=}& \overrightarrow{\mathbf{F}}\mathbf{.}\overrightarrow{\mathbf{d}}\end{array}$

In both cases, there is no acceleration, so the lifting force is equal to the weight of the object.

(a) The work done on the roof:

Calculate the work done on the roof by the forces making the lift as follow.

$\begin{array}{rcl}{\mathrm{W}}_1& =& \overrightarrow{{\mathrm{F}}_1}.\overrightarrow{{\mathrm{d}}_1}\\ & =& \left(360\mathrm{kN}\right)\left(0.1\mathrm{m}\right)\\ & =& 36\mathrm{kJ}\end{array}$

Hence, the work done on the roof by the forces making the lift is 36 kJ.

(b) The work done on the car:

Determine the work done on the car as follow.

$\begin{array}{rcl}{\mathrm{W}}_2& =& {\overrightarrow{\mathrm{F}}}_2.\overrightarrow{{\mathrm{d}}_2}\\ & =& \left(4000\mathrm{N}\right)\left(0.050\mathrm{m}\right)\\ & =& 2\times {10}^2\mathrm{J}\end{array}$

Hence, the work done on the car is $2\times {10}^2\mathrm{J}$.