Answer:
17.3 m/s (1 d.p.)
Explanation:
Kinetic Energy
[tex]E_k=\sf \dfrac{1}{2}mv^2[/tex]
where:
- [tex]E_k[/tex] = kinetic energy in joules (J)
- m = mass in kilograms (kg)
- v = speed in meters per second (m/s)
Gravitational potential energy
[tex]E_p=\sf mgh[/tex]
where:
- [tex]E_p[/tex] = gravitational potential energy in joules (J)
- m = mass in kilograms (kg)
- g = gravitational field strength in newtons per kilogram (N/kg)
- h = change in height in meters (m)
Principle of Conservation of Energy
Gravitational potential energy at the top = kinetic energy at the bottom
[tex]\implies \large \text{$ E_{p \sf (top)}=E_{k \sf(bottom)} $}[/tex]
Given:
The gravitational field strength of the Earth is 10 N/kg (10 newtons per kilogram), therefore:
Substituting the values into the formula and solving for v:
[tex]\implies \large \text{$ E_{p \sf (top)}=E_{k \sf(bottom)} $}[/tex]
[tex]\implies \sf mgh=\dfrac{1}{2}mv^2[/tex]
[tex]\implies \sf (5.0)(10)(15)=\dfrac{1}{2}(5.0)v^2[/tex]
[tex]\implies \sf 750=\dfrac{5}{2}v^2[/tex]
[tex]\implies \sf v^2=300[/tex]
[tex]\implies \sf v=\sqrt{300}[/tex]
[tex]\implies \sf v=17.3\:\: m/s \:\: (1\:d.p.)[/tex]
