Ọrụ Net Ike ike ndọda Ike Kinetic - Nsogbu na Ngwọta

4 Ọrụ Net Ike ike ndọda Ike Kinetic - Nsogbu na Ngwọta

1. Ihe dị kilogram ise n'ogologo mita iri n'elu ala. Ọsọ ọsọ n'ihi ike ndọda bụ 10 m/s². Gịnị bụ ọrụ e mere ihe ahụ iji bulie ya elu ruo n'ịdị elu nke mita iri na ise n'elu ala?

A maara:

Mass nke ihe (m) = 5 kg

Ogologo (h) = mita 10

Ọsọ ọsọ n'ihi ike ndọda (g) = 10 m/s²

Chọrọ: E tinyere ọrụ n'ihe ahụ iji bulie ya elu ruo n'ogo nke mita iri na ise n'elu ala.

Ngwọta:

W = F d = wh = mgh

W = work, F = force, d = nnyefe, w = ibu ibu, h = height, m = mass, g = acceleration due to gravity

Work done on the object :

W = m g h = (5 kg)(10 m/s²)(10 m) = 500 kg m²/s² = 500 Joule

2. A 2-kg object moves along smooth horizontal surface with the speed of 2 m/s. If the final speed of the object is 5 m/s. What is the work done on the object.

A maara:

Ibu ihe (m) = 2 kg

Ọsọ mbụ (v)_o) = 2 m/s

Ọsọ ikpeazụ (v)_t) = 5 m/s

Chọrọ: Work done on the object

Ngwọta:

The work-kinetic energy principle states that the net work done on an object is equal to the change in the object’s ike.

W = ΔKE = KE₂ – KE₁ = 1/2 mv_t² – 1/2 mv_o² = 1/2 m (v_t² - v_o²) = 1/2 (2)(5)²-2²) = (1)(25-4) = 21 Joule

3. An object with mass of 10-kg initially at rest on a smooth floor. After pushed in 3 seconds, the object accelerated at 2 m/s². What is the work done on the object.

A maara:

Ibu (m) = 10 kg

Ọsọ mbụ (v)_o) = 0 (initially at rest)

Oge etiti oge (t) = sekọnd 3

Ọsọ ọsọ (a) = 2 m/s²

A chọrọ: Ọrụ (W)

Ngwọta:

First, determine distance using the equation of the mgbe niile osooso ngagharị.

d = v_o t + ½ na² = (0)(3) + ½ (2)(3)²

d = 0 + (1)(9) = 9

Work done on the object :

W = F s = (m a)(d) = (10)(2)(9) = 180 Joule

4. A block with mass of 1.5 kg accelerated upward by a constant force F = 15 N on a inclined plane as shown in figure below. Acceleration due to gravity is 10 m/s² and no friction between block and ụgbọelu chọrọ. What is the net work done on the object.

A maara:

Ibu nke blọk (m) = 1.5 kg

Ọsọ ọsọ n'ihi ike ndọda (g) = 10 m/s²

Weight of block (w) = m g = (1.5)(10) = 15 Newton

Ike (F) = 15 Newton

Chọrọ: Net work is done on the object

Ngwọta:

There are two forces do work on the block. First, a force parallel to the block’s displacement that is Force F. Second, horizontal component of the weight of block (w_x).

Work done by force (F) :

W₁ = F s = (15 N)(2 m) = 30 N m = 30 Joules

Work done by horizontal component of weight (w_x):

W₂ = w_x s = (w mmehie 30^o)(2 m) = (15 N)(1/2)(2 m) = 15 N m = 15 Joules

Net work done on the object :

W_net = W₁ - W₂ = 30 J – 15 J = 15 Joule

1. What is the work-energy theorem in terms of net work and kinetic energy?
   • Azịza: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. Mathematically: WnetNa=ΔKE.
2. If you lift an object to a certain height and then hold it there, how much work are you doing on the object while holding it stationary?
   • Azịza: While holding the object stationary, you are doing zero work on it because there is no displacement. Work is defined as the force times the displacement in the direction of the force, so no displacement means no work.
3. What is the gravitational potential energy of an object at ground level?
   • Azịza: The gravitational potential energy of an object at ground level is typically taken to be zero. Gravitational potential energy is relative, and ground level is a common reference point.
4. How does the kinetic energy of an object change if its velocity is doubled?
   • Azịza: Kinetic energy is proportional to the square of velocity. If the velocity of an object is doubled, its kinetic energy will increase by a factor of four.
5. When is the gravitational potential energy of a falling object maximum?
   • Azịza: The gravitational potential energy of a falling object is maximum just before it starts falling, i.e., when it is at its highest point above the ground (or reference point).
6. In the absence of air resistance, what can be said about the relationship between the potential energy lost and the kinetic energy gained by a freely falling object?
   • Azịza: In the absence of air resistance, the potential energy lost by a freely falling object will be equal to the kinetic energy it gains. This is a direct consequence of the conservation of mechanical energy.
7. What happens to the gravitational potential energy of an object if it is raised to twice its original height above the ground?
   • Azịza: Gravitational potential energy is proportional to height. If an object is raised to twice its original height, its gravitational potential energy will double.
8. When is the net work done on an object zero?
   • Azịza: The net work done on an object is zero when the total work done by all forces acting on it sum up to zero, or equivalently, when there’s no change in the object’s kinetic energy.
9. If an object is moving at a constant velocity, what can be inferred about the net work done on it over a certain distance?
   • Azịza: If an object is moving at a constant velocity, it means there’s no acceleration, and hence no net force acting on it. Therefore, the net work done on the object over any distance will be zero.

10. Why does an object at a higher altitude have more gravitational potential energy even though it’s further from the Earth’s center?

    • Azịza: Gravitational potential energy is based on an object’s position relative to a reference point (often the Earth’s surface) and the force of gravity acting on it. An object at a higher altitude is higher above this reference point, which means it has been moved against the gravitational force to that position. As a result, it has stored energy due to its position, and this stored energy is its gravitational potential energy.