Question

A net force of 70 N accélérâtes a 3-kg mass over a distance of 12 m A. What is the work done by the net force.B. What is the increase in kinetic energy of the mass?

Answer 1

A. 840 J

B. 840 J

Explanation:

Given:

Force (F) = 70 N

Mass (m) = 3 kg

Distance (d) = 12 m

A.

The work is given by the following equation:

`\begin{gathered} W=F\cdot d \\ \\ W=70\cdot12 \\ \\ W=840\text{ J} \end{gathered}`

Work is equal to 840 joules

B.

Initial speed = 0

Final speed = v

We calculate the acceleration just like this:

`\begin{gathered} F=m\cdot a \\ \\ a=(F)/(m) \\ \\ a=(70)/(3)=23.33\text{ m/s}^2 \end{gathered}`

Knowing this we can calculate the speed:

`\begin{gathered} v^2=u^2+2ad \\ \\ v^2=0^2+2\cdot23.33\cdot12 \\ \\ v^2=559.92 \\ \\ v=√(559.92) \\ \\ v=23.66\text{ m/s} \end{gathered}`

Now we calculate the increase in kinetic energy:

`\begin{gathered} KE=(1)/(2)m(v^2-u^2) \\ \\ KE=(1)/(2)\cdot3\cdot(23.66^2-0^2) \\ \\ KE=839.69\cong840\text{ J} \end{gathered}`

Therefore, the increase in kinetic energy is 840 joules.