Question

Austin does his Power lifting every morning to stay in shape. He lifts a 90 kg barbell, 2.3 m above the ground.a) How much energy does it have when it was on the ground? Jb)How much energy does it have after being lifted 2.3 m? Jc) What kind of energy does it have after being lifted? d) How much work did Austin do to lift the barbell? Je) If he lifted it in 1.9s, what was his power? W

Answer 1

`\begin{gathered} (a)0J \\ (b)2030J \\ (c)\text{ Potential energy} \\ (d)2030J \\ (e)1070W \end{gathered}`

Step-by-step explanation

Parameters given:

Mass of barbell, m = 90 kg

Height above ground, h = 2.3 m

(a) We want to find the energy the barbell has on the ground. #

When it is on the ground, the barbell is stationary, which means its velocity is 0 m/s, hence, its kinetic energy is also 0 J, since kinetic energy is given as:

`\begin{gathered} KE=(1)/(2)mv^2 \\ KE=(1)/(2)\cdot m\cdot0=0J \end{gathered}`

Also, on the ground, it is at a height of 0 m, hence, its potential energy is 0 J:

`\begin{gathered} PE=mgh \\ PE=m\cdot g\cdot0=0J \end{gathered}`

where g = acceleration due to gravity

Therefore, on the ground, the energy the barbell had was 0 J.

(b) After it had been lifted 2.3 m, its height above the ground became 2.3 m.

Now, we can find the potential energy possessed by the barbell:

`\begin{gathered} PE=90\cdot9.8\cdot2.3 \\ PE=2028.6J\approx2030J \end{gathered}`

After it is lifted, it is once again stationary, hence, it has no kinetic energy.

Therefore, the energy the barbell has after it has been lifted 2.3 m is 2070J.

(c) As stated in (b) above, after being lifted, the barbell only possesses potential energy since it is at a height above the ground and it is not moving.

(d) The work done in lifting the barbell is equal to the force applied multiplied by the height moved by the barbell.

That is:

`W=F\cdot d`

The force applied is equal to the weight of the barbell:

`\begin{gathered} F=W=mg \\ F=90\cdot9.8 \\ F=882N \end{gathered}`

Therefore, the work done is:

`\begin{gathered} W=882\cdot2.3 \\ W=2028.6J\approx2030J \end{gathered}`

(e) He lifted the barbell in 1.9 seconds. To find his power, we have to divide the work done by the time taken to do the work.

That is:

`\begin{gathered} P=(W)/(t) \\ P=(2030)/(1.9) \\ P=1068.4W\approx1070W \end{gathered}`

That was his power.