3. State Newton's first and second laws of equation. A. A body of mass 2kg move vertically upwards has its velocity increased uniformly from 15m/s to 45m/s in 5s. neglecting air resistance, the upward vertical force acting on the body. [g=10ms 2] B. The velocity ratio and efficiency of pulley are 8 and 85% respectively. How much effort is required to lift a load of 120kg with this system [g=10m/s²]

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03-07-2022
Physics

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3. State Newton's first and second laws of equation. A. A body of mass 2kg move vertically upwards has its velocity increased uniformly from 15m/s to 45m/s in 5s. neglecting air resistance, the upward vertical force acting on the body. [g=10ms 2] B. The velocity ratio and efficiency of pulley are 8 and 85% respectively. How much effort is required to lift a load of 120kg with this system [g=10m/s²]

Sagot :

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rspill6 rspill6 · Answer 1 · 2022-07-03T10:03:43-04:00

Answer:

See below.

Explanation:

Newton's First Law

A body at rest persists in its state of rest, and a body in motion remains in constant motion along a straight line unless acted upon by an external force.

Newton's Second Law

A body's acceleration is directly proportional to the force exerted on it and is in the same direction as the force. F=mA (Force = Mass x Acceleration)

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A. A body of mass 2kg move vertically upwards has its velocity increased uniformly from 15m/s to 45m/s in 5s. neglecting air resistance, the upward vertical force acting on the body. [g=10ms 2] B. The velocity ratio and efficiency of pulley are 8 and 85% respectively. How much effort is required to lift a load of 120kg with this system [g=10m/s²]

Force = Mass x Acceleration

Acceleration is (45m/s - 15 m/s)/5 sec) or 6 m/sec^2

Force = 2kg*(6 m/sec^2)

Force = 12 kg*m/s^2 1 kg*m/s^2 is the SI unit for force, and is named 1 Newton(N), after . . . Newton.

Force = 12N

B. I'm not sure how to interpret "The velocity ratio and efficiency of pulley are 8 and 85% respectively." I'll assume "effort" is the energy required.

I'll calculate the energy required in a perfect system, and then we'll consider the efficiency issues.

By lifting an object with mass off the Earth's surface (height = 0 meters) we impart a potential energy, PE, to the object. This energy is stored in the object until it is allowed to move back closer to Earth. This potential energy is given by the equation:

PE = mgh

where PE is potential energy in Joules, m is mass in kg, g is the acceleration due to Earth's gravity (9.8 m/sec^2, but we'll use 10 m/sec^2 as stated in the problem), and h is the vertical distance moved, in meters.

PE = (120kg)(10 m/sec^2)(h). We aren't given a height, h. Let's express the answer in terms of 1 meter: How much potential energy is put into the 120kg load to raise it 1 meter?

PE = (120kg)(10 m/sec^2)(1 m)

PE = 1200 kg*m^2/sec^2 or 1200 kg*m/sec^2 per meter

The unit kg*m/sec^2 is 1 Joule

PE = 1200 Joules/meter

1200 Joules are required to raise 120kg 1 meter. This is increased due to the less-tha-100% efficiency of 85%.

At 85% efficiency, the energy required would be 1200J/0.85 = 1412 J to raise 120kg 1 meter, with 85% efficiency, I don't6 know how to handle the velocity ratio number.

Sagot :

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