Question

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider dives to a height of 278 meters. Determine the gliders new speed.

Answer 1

Answer:

The glider's new speed is 68.90 m/s

Explanation:

Principle Of Conservation Of Mechanical Energy

The mechanical energy of a system is the sum of its kinetic and potential energy. When the only potential energy considered in the system is related to the height of an object, then it's called the gravitational potential energy. The kinetic energy of an object of mass m and speed v is

$\displaystyle K=\frac{1}{2}mv^2$

The gravitational potential energy when it's at a height h from the zero reference is

$U=mgh$

The total mechanical energy is

$M=K+U$

$\displaystyle M=\frac{1}{2}mv^2+mgh$

The principle of conservation of mechanical energy states the total energy is constant while no external force is applied to the system. One example of a non-conservative system happens when friction is considered since part of the energy is lost in its thermal manifestation.

The initial conditions of the problem state that our glider is glides at 416 meters with a speed of 45.2 m/s. The initial mechanical energy is

$\displaystyle M_1=\frac{1}{2}m(45.2)v_o^2+m(9.8)(416)$

Operating in terms of m

$\displaystyle M_1=1021.52m+4076.8m$

$\displaystyle M_1=5098.32m$

Then we know the glider dives to 278 meters and we need to know their final speed, let's call it $v_f$. The final mechanical energy is

$\displaystyle M_2=\frac{1}{2}mv_f^2+m(9.8)(278)$

Operating and factoring

$\displaystyle M_2=m(\frac{1}{2}v_f^2+2724.4)$

Both mechanical energies must be the same, so

$\displaystyle m(\frac{1}{2}v_f^2+2724.4)=5098.32m$

Simplifying by m and rearranging

$\displaystyle \frac{v_f^2}{2}=5098.32-2724.4$

Computing

$v_f=\sqrt{4747.84}=68.90\ m/s$

The glider's new speed is 68.90 m/s