Answer to Question #90435 in Mechanics | Relativity for AWESOME

Question

A body of mass 5kg is released from a height of 20m. calculate its velocity at the ground and when it is 10m from the ground. What is the average resistant force exerted by the sand when the body penetrates a distance of 3/4m

Answer 1

To find velocity of the body we use the equation

$${{v}^{2}}=v_{0}^{2}-2g\left( y-{{y}_{0}} \right),$$

where ${{v}_{0}}$ is the initial velocity of the body, ${{v}_{0}}=0$ ,

${{y}_{0}}$ is the initial height, ${{y}_{0}}=20\,m$ ,

y is the height at which we need to find the speed of the body, $y=0\,m$ at the ground, or $y=10\,m$ at a height of 10 m from the ground. 

 g is the acceleration of gravity, $g=10\,m\cdot {{s}^{-2}}$ .

Substituting the known values, we get:

1) at the ground

$$v=\sqrt{-2\cdot 10\,m\cdot {{s}^{-2}}\cdot \left( 0-20 \right)m}=20\,m/s$$

2) at a height of 10 m from the ground

$$v=\sqrt{-2\cdot 10\,m\cdot {{s}^{-2}}\cdot \left( 10-20 \right)m}$$

$$=10\sqrt{2}\,m/s\approx 14.14\,m/s$$

To find the average resistant force we use the work-energy theorem: The net work ${{W}_{net}}$  on a system equals the change in the quantity of kinetic energy

$${{W}_{K}}=\frac{1}{2}m{{v}^{2}}$$

Calculate kinetic energy of the body near the ground. Substitute $m=5\,kg$ and$v=20\,m/s$

$${{W}_{K}}=\frac{1}{2}\cdot 5\,kg\cdot {{\left( 20\,m/s \right)}^{2}}=1000\,J=1\,kJ.$$

The net work of an average friction force f on a system is

$${{W}_{net}}=f\cdot d$$

where d is the distance it takes to stop, d=3/4m. Equate the kinetic energy and the work of an average friction force

$${{W}_{net}}={{W}_{K}}$$

or

$$f\cdot d=1\,kJ$$

Then we get an average friction force

$$f=\frac{1\,kJ}{3/4\,m}=\frac{4}{3}\,\frac{kJ}{m}\approx 1.33\,kN$$