Question

A 25 kg bear slides, from rest, 12 m down a lodge pole pine tree, moving with a speed of 5.6 m/s just before hitting the ground. (a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?

Short answer

a) The change in gravitational energy of the bear-earth system is -2.9 kJ.

b) The kinetic energy of the bear just before hitting the ground is $3.9\times {10}^2\mathrm{J}$.

c) The average frictional force acting on the bear is $2.1\times {10}^2\mathrm{N}$.

Step-by-step solution

Step 1: Given Data

The mass of the bear is 25 kg.

The distance travelled by bear is 12 m.

The speed of the bear just before hitting the ground is 5.6 m/s.

Determining the concept

Total energy of the bear – earth system is conserved. The potential energy at the top of the tree is converted to kinetic energy as it reaches the bottom, and some amount of energy is lost during the motion due to friction.

Formulae are as follow:

Potential energy,$PE=mgh$

Kinetic energy,$KE=\frac{1}{2}m{v}^2$

$\mathrm{Frictional}\mathrm{force}=\mathrm{Energy}\mathrm{lost}/\mathrm{distance}\mathrm{travelled}$

Step 3(a): Determining the change in gravitational energy of the bear-earth system

The potential energy of the bear at the top of the tree is given by,

$$\begin{gathered} PE=U_f-U_i \\ =-mgh \\ =-25\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 12\mathrm{m} \\ =-2.9\times {10}^3\mathrm{J} \\ =-2.9\mathrm{kJ} \end{gathered}$$

Hence, the change in gravitational energy of the bear-earth system is -2.9 kJ.

Step 4(b): Determining the kinetic energy of the bear just before hitting the ground

Thekinetic energy of the bear just before hitting the ground is given by,

$$\begin{gathered} KE=\frac{1}{2}m{v}^2 \\ =\frac{1}{2}\times 25\mathrm{kg}\times {\left(5.6\mathrm{m}/\mathrm{s}\right)}^2 \\ =3.9\times {10}^2\mathrm{J} \end{gathered}$$

Hence, the kinetic energy of the bear just before hitting the ground is $3.9\times {10}^2\mathrm{J}$.

Step 5(c): Determining the average frictional force acting on the bear

Due to frictional force acting on the bear, some of its initial PE is lost during its travel to the bottom of the tree. Thus, the energy lost during the travel is calculated as,

$$\begin{gathered} ∆E=PE\left(\mathrm{at}\mathrm{the}\mathrm{top}\right)-KE\left(\mathrm{at}\mathrm{the}\mathrm{bottom}\right) \\ ∆\mathrm{E}=2.9\times {10}^3-3.9\times {10}^2 \\ =2510\mathrm{J} \end{gathered}$$

This energy loss is related to the frictional force as,

$$\begin{gathered} ∆E=F_{f}d∆F_f \\ =\frac{∆E}{d} \\ =\frac{2510}{12} \\ =2.10\times {10}^2\mathrm{N} \end{gathered}$$

Hence, the energy loss is related to the frictional forceis$2.10\times {10}^2\mathrm{N}$.

Therefore, the law of conservation of energy holds true for the bear-earth system. The total energy at the top of the tree is potential energy, some part of which is converted to kinetic energy at the bottom of the tree, and some part is lost due to the frictional force acting on the bear.