What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?
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The question is missing its alternatives. Here is the complete question.
Consider two identical objects released from rest high above the surface of Earth. In Case 1, the object is released from a height above the surface of Earth equal to 1 Earth radius, and we measure its kinetic energy just before it hits the Earth to be K1. In Case 2, the obejct is released from a height above the surface of Earth equal to 2 Earth radii and its kinetic energy just before it hits is K2.
1. Compare the kinetic energy of the two objects just before they hit the surface of the earth.
a) K2 = 2K1; b) K2 = 4K; c) K2 = (4/3)K1; d) K2 = (3/2)K1;
Answer: C) K2 = (4/3)K1
Explanation: As it is related to the gravity of the Earth, the potencial energy is: U(r)= - + U₀
In this case, U₀=0, G is the universal gravitational constant, Me is the mass of Earth, m is the mass of the object and r is the distance between the center of the Earth and the object.
The potencial energy of an object of mass m on the surface of the Earth is:
Usurface = -
The potencial energy of the object in Case 1 is
U1 = -
For the Case 2:
U2 = -
The potencial change in Case 1:
ΔU1 = - G.Me.m.() = -
For Case 2:
ΔU2 = - G.Me.m() = -
Comparing ΔK1 and ΔK2 equals comparing ΔU1 and ΔU2:
Δ = (-2/3)(-1/2) = 4/3
So, comparing kinetic energies, K2 is 4/3 of K1.