Asked by Anonymous
initially a 10 kg block is sliding along a level floor at a speed of 5 m/s. the coefficient between the block and floor is exactly 0.5?
a) what are the kinetic, potentially and total energy initially . KE=125 J but what is the PE
b) what is the force fo friction done by floor on block?
c. what is the work done by t he frictional face on the block slides 2 m across the floor?
d. what are the final kinetic, Potential and total energies?
3. what is the blocks speed after sliding 2 m?
a) what are the kinetic, potentially and total energy initially . KE=125 J but what is the PE
b) what is the force fo friction done by floor on block?
c. what is the work done by t he frictional face on the block slides 2 m across the floor?
d. what are the final kinetic, Potential and total energies?
3. what is the blocks speed after sliding 2 m?
Answers
Answered by
Anonymous
If you define height as zero at the floor, the PE is 0 at the floor.
a. kinetic = (1/2) m v^2 =.5 * 10 * 25 = 125 J = total
b.force = mu m g = 0.5 * 10 * 9.81 = 49.05 Newtons
c. F * distance in direction of force = 49.05 * 2 = 98.1 Joules
d potential is still zero, Ke = 125 - 98.1 = total
e. (1/2) m v^2 = total
a. kinetic = (1/2) m v^2 =.5 * 10 * 25 = 125 J = total
b.force = mu m g = 0.5 * 10 * 9.81 = 49.05 Newtons
c. F * distance in direction of force = 49.05 * 2 = 98.1 Joules
d potential is still zero, Ke = 125 - 98.1 = total
e. (1/2) m v^2 = total
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