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Sagot :
a) The weight of a person is given by:
[tex]W=mg[/tex]
where m is the mass of the person and g is the gravitational acceleration. In this problem, the mass of the person is m=60.0 kg, while the value of g is [tex]g=9.81 m/s^2[/tex], therefore the weight of the person is
[tex]W=(60 kg)(9.81 m/s^2)=588.6 N[/tex]
b) The gravitational potential energy at the top of the hill is U=1000 J. The potential energy is also given by the product between the weight W and the height of the hill h:
[tex]U=Wh[/tex]
If we rearrange the equation, we can calculate the height of the hill, h:
[tex]h= \frac{U}{W} = \frac{1000 J}{588.6 N} =1.70 m[/tex]
c) At the bottom of the hill, all the potential energy at the top of the hill has converted into kinetic energy:
[tex]U=K[/tex]
[tex]U= \frac{1}{2} mv^2[/tex]
where m is the mass and v is the speed. If we rearrange the formula, we can calculate the speed at the bottom of the hill:
[tex]v=\sqrt{\frac{2U}{m}}=\sqrt{\frac{2\cdot 1000 J}{60.0 kg}}=5.8 m/s[/tex]
[tex]W=mg[/tex]
where m is the mass of the person and g is the gravitational acceleration. In this problem, the mass of the person is m=60.0 kg, while the value of g is [tex]g=9.81 m/s^2[/tex], therefore the weight of the person is
[tex]W=(60 kg)(9.81 m/s^2)=588.6 N[/tex]
b) The gravitational potential energy at the top of the hill is U=1000 J. The potential energy is also given by the product between the weight W and the height of the hill h:
[tex]U=Wh[/tex]
If we rearrange the equation, we can calculate the height of the hill, h:
[tex]h= \frac{U}{W} = \frac{1000 J}{588.6 N} =1.70 m[/tex]
c) At the bottom of the hill, all the potential energy at the top of the hill has converted into kinetic energy:
[tex]U=K[/tex]
[tex]U= \frac{1}{2} mv^2[/tex]
where m is the mass and v is the speed. If we rearrange the formula, we can calculate the speed at the bottom of the hill:
[tex]v=\sqrt{\frac{2U}{m}}=\sqrt{\frac{2\cdot 1000 J}{60.0 kg}}=5.8 m/s[/tex]
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