Respuesta :
Answer:
Part a)
[tex]W = 1200.7 J[/tex]
Part B)
[tex]KE = 1200.7 J[/tex]
Part C)
[tex]v = 21.7 m/s[/tex]
Part D)
no change in answer of part A) if there is air resistance in that problem.
Explanation:
Part A)
Work done by the gravity is given as
[tex]W = mgH[/tex]
now we have
m = 5.10 kg
H = 24 m
now we will have
[tex]W = (5.10 kg)(9.81 m/s^2)(24 m)[/tex]
[tex]W = 1200.7 J[/tex]
Part B)
As per work energy theorem we can say that work done by all forces = change in kinetic energy of the system
So we will have
[tex]W = KE[/tex]
[tex]KE = 1200.7 J[/tex]
Part C)
Now we know that the kinetic energy is given as
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]1200.7 = \frac{1}{2}(5.10) v^2[/tex]
[tex]v = 21.7 m/s[/tex]
Part D)
Work done by gravity only depends on force due to gravity and its displacement so there will be no change in answer of part A) if there is air resistance in that problem.
Work done is the change in the energy of the system. The kinetic energy of the watermelon is 1200.744 J.
What is the work done?
Work done can be defined as the change in the energy of the object.
A.) We know that work done by the gravity on the watermelon is the change in the potential energy of the watermelon, therefore,
Work done due to gravity = change in the potential energy of the system
[tex]W = \triangle PE\\W = mg(h_1-h_0)\\\\[/tex]
Since the mass of the watermelon is 5.10 kg while the height from which the watermelon is dropped is 24 m, thus, the work done is equal to,
[tex]W = 5.10 \times 9.81 \times (24-0)\\W = 1200.744\rm\ J[/tex]
B.) We know that the potential energy of the watermelon at the height will be completely converted to the kinetic energy of the watermelon as its hits the ground, therefore, the work done will be changed in the kinetic energy of the watermelon, also, the kinetic energy of the watermelon at the height of 24 m will be zero.
[tex]W = \triangle KE\\1200.744 = KE[/tex]
C.) We know the kinetic energy of the watermelon, also the mass of the watermelon. therefore, the velocity of the watermelon just before it hits the ground will be,
[tex]KE =\dfrac{1}{2}mv^2\\\\1200.744 = \dfrac{1}{2}\times 5.10 \times v^2\\\\v = 21.7\ m/s[/tex]
D.) Work done by gravity only depends on the force due to gravity and its displacement so there will be no change in the result of part A if there is air resistance in that problem.
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