1. Mechanical energy of the system: 3597 J
The mechanical energy of the system is equal to the sum of the kinetic energy (K) and the gravitational potential energy (U):
[tex]E=K+U[/tex]
However, the sled starts from rest, so the initial kinetic energy is zero: K=0. Therefore, at the top of the hill all the mechanical energy is just potential energy, given by:
[tex]U=mgh[/tex]
where m=50.0 kg is the mass, g=9.8 m/s^2 the acceleration due to gravity and h=7.34 m is the height. Substituting,
[tex]E=U=(50.0 kg)(9.8 m/s^2)(7.34 m)=3597 J[/tex]
2. Sleed's speed at the bottom of the hill: 12.0 m/s
At the bottom of the hill, h=0, so the potential energy has become zero: U=0. However, the mechanical energy must be conserved, so all this energy has been converted into kinetic energy:
[tex]E=K=3597 J[/tex]
The formula for the kinetic energy is:
[tex]K=\frac{1}{2}mv^2[/tex]
where m=50.0 kg is the mass and v the speed of the sled at the bottom of the hill. Re-arranging the equation, we find v:
[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3597 J)}{50.0 kg}}=12.0 m/s[/tex]
The mechanical energy of the system is 3596.6 J and the speed of sled at the bottom of hill is 11.99 m/s.
Given data:
The mass of child-sled system is, m = 50.0 kg.
The height of system on the hill is, h = 7.34 m.
The mechanical energy of the system at any point is the combination of kinetic energy as well as potential energy of the system at that point.
Mechanical Energy = Kinetic Energy + Potential Energy
[tex]ME =\dfrac{1}{2}mv^{2} +mgh[/tex]
Here, v is the speed of system, since child and sled were at rest initially. Then, v = 0 m/s.
Solving as,
[tex]ME =\dfrac{1}{2} \times 50 \times 0^{2} +(50 \times 9.8 \times 7.34)\\\\ME = 3596.6 \;\rm J[/tex]
And at the bottom of the hill, the potential energy of the system is zero. Which means,
Mechanical Energy = Kinetic Energy + Potential Energy
Mechanical Energy = Kinetic Energy + 0
[tex]ME = \dfrac{1}{2}mv'^{2}[/tex]
Here, v' is the speed of sled at bottom of hill.
Solving as,
[tex]3596.6 = \dfrac{1}{2} \times 50 \times v'^{2}\\\\v' = 11.99 \;\rm m/s[/tex]
Thus, we can conclude that the mechanical energy of the system is 3596.6 J and the speed of sled at the bottom of hill is 11.99 m/s.
Learn more about the mechanical energy here;
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