Question

A child pushes a $2.2\text{kg}$ swing that is initially at rest. The net force on the swing over time is shown below.


What is the swing's speed at $t=40\text{ms}$? Round off the answer to two decimal places.

• A. $0.36\text{m/s}$
• B. $0.48\text{m/s}$
• C. $0.15\text{m/s}$
• D. $0.26\text{m/s}$

Answer 1

The correct option is A $0.36\text{m/s}$

Area under force vs. time graph gives change in momentum $\Delta P$.
$\int F.dt=\Delta P=P_f-P_i$
Area under graph $=\frac{1}{2}\times \text{base}\times \text{height}$
$=\frac{1}{2}\times 40\times {10}^{-3}\times 40$
$\therefore \Delta P=\frac{1}{2}\left(0.040\text{s}\right)\left(40\text{N}\right)=0.80\text{kg m/s}$
We can use the change in momentum to solve for the final speed,
$\Delta P=m\Delta v$
$\Delta v=\frac{\Delta P}{m}$
$\Rightarrow v_f-0=\frac{\left(0.80\frac{\text{kg m}}{\text{s}}\right)}{2.2\text{kg}}$
$\therefore v_f\approx 0.363\text{m/s}$
Rounded off to two significant figures, the final speed of the swing is $0.36\text{m/s}$