Question

A60.0 kgcircus performer slides4.00 mdown a pole to the circus floor, starting from rest.

What is the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole (a) is negligible (she will be hurt) and (b) has a magnitude of 500 N ?

Short answer

Kinetic energy when,

1. Friction force is negligible $K{E}_f=2.35\times {10}^3\mathrm{J}$
2. Friction force $F_f=500N$ is $K{E}_f=352\mathrm{J}$.

Step-by-step solution

Given data:

Mass of circus performer, m=60 kg

Distance down the pole, h=4.0 m

Acceleration due to gravity, $g=9.8m/s^2$

To understand the concept:

The law of conservation of energy states that energy cannot be created or destroyed - only converted from one form of energy to another. This means that the system always has the same amount of energy unless it is added from outside.

According to the law of conservation of energy, total mechanical energy is constant.

Formula:

$InitialTotalenergy=Finaltotalenergy$

(a) The kinetic energy of the performer if the frictional force is negligible:

Calculate the kinetic energy of the performer as she reaches the floor if the frictional force on her from the pole is negligible:

If friction is negligible,

According to law of conservation of energy,

$KEatthefloor=PEattheheight$

Hence,

$$\begin{gathered} KE=mgh \\ =60\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 4\mathrm{m} \\ =2.35\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the required kinetic energy is $2.35\times {10}^3\mathrm{J}$.

(b) The kinetic energy of the performer if the friction has a magnitude of  :

Calculate the kinetic energy of the performer as she reaches the floor if the friction has a magnitude of 500 N.

Therefore,

The frictional force, $F_f=500N$

If friction is available,

According to law of conservation of energy,

$$\begin{gathered} KE=PE+Workdone \\ =mgh-F_{f}h \end{gathered}$$

Negative work is done because friction force is in opposite direction.

$$\begin{gathered} KE=\left[\left(60\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\times \left(4\mathrm{m}\right)\right]-\left(500\mathrm{N}\times 4\mathrm{m}\right) \\ =2352\mathrm{J}-2000\mathrm{J} \\ =352\mathrm{J} \end{gathered}$$

Hence, the required kinetic energy is 352 J.