To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
7.2 as used in the equation
Answer:
This means that the kinetic energy of second object is 48times that of the first object
Explanation:
Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,
Kinetic energy = 1/2mv² where;
m is the mass of the object
v is the velocity of the object
If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;
K1 = 1/2mv²
For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;
K2 = 1/2(3m)(4v)²
K2 = 1/2(3m)(16v²)
K2 = (3m)(8v²)
K2 = 24mv²
To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;
K2/K1 = 24mv²/(1/2)mv²
K2/K1 = 24/(1/2)
K2/K1 = 48
K2 = 48K1
This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.
