Question

A lab cart with a mass of 530 grams has an initial speed of 1.30 meters per second. While the cart rolls through a distance of 34.0 centimeters, a constant force of 4.29 newtons pushes the cart in the direction it's moving. At the same time, friction does -0.237 joules of work. What is the cart's final speed? Include units in your answer. Answer must be in 3 significant digits.

Answer 1

The net work done on the cart can be given as,

`W=Fd+W_f`

The change in kinetic energy of the cart is,

`\Delta K=(1)/(2)m(v^2-u^2_{})`

According to work energy theorem,

`W=\Delta K`

Substitute the known expression,

`\begin{gathered} Fd+W_f=(1)/(2)m(v^2-u^2) \\ v^2-u^2=(2(Fd+W_f))/(m) \\ v^2=(2(Fd+W_f))/(m)+u^2 \end{gathered}`

Substitute the known values,

`\begin{gathered} v^2=\frac{2((4.29\text{ N)(34.0 cm)(}\frac{1\text{ m}}{100\text{ cm}})\text{-0.237 J)}}{(530\text{ g)(}\frac{1\text{ kg}}{1000\text{ g}})}+(1.30m/s)^2 \\ =4.610m^2s^(-2)+1.69m^2s^(-2) \\ v=\sqrt[]{6.300m^2s^(-2)} \\ \approx2.51\text{ m/s} \end{gathered}`

Thus, the final speed of cart is 2.51 m/s.