**4 Net work Gravitational potential energy Kinetic energy – Problems and Solutions**

1. A 5-kg object at the height of 10-meters above the ground. Acceleration due to gravity is 10 m/s^{2}. What is the work done on the object to moves it upward to the height of 15-meters above the ground?

__Known :__

Mass of object (m) = 5 kg

Height (h) = 10 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted:__ Work was done on the object to moves it upward to the height of 15-meters above the ground.

__Solution :__

W = F d = w h = m g h

*W = work, F = force, d = displacement, w = weight, h = height, m = mass, g = acceleration due to gravity*

Work done on the object :

W = m g h = (5 kg)(10 m/s^{2})(10 m) = 500 kg m^{2}/s^{2} = 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.

__Known :__

Mass of object (m) = 2 kg

Initial velocity (v_{o}) = 2 m/s

Final velocity (v_{t}) = 5 m/s

__Wanted:__ Work done on the object

__Solution :__

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

W = ΔKE = KE_{2} – KE_{1} = 1/2 m v_{t}^{2} – 1/2 m v_{o}^{2 }= 1/2 m (v_{t}^{2} – v_{o}^{2}) = 1/2 (2)(5^{2}-2^{2}) = (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/s^{2}. What is the work done on the object.

__Known :__

Mass (m) = 10 kg

Initial velocity (v_{o}) = 0 (initially at rest)

Time interval (t) = 3 seconds

Acceleration (a) = 2 m/s^{2}

__Wanted :__ Work (W)

__Solution :__

First, determine distance using the equation of the constant acceleration motion.

d = v_{o} t + ½ a t^{2 }= (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/s^{2} and no friction between block and inclined plane. What is the net work done on the object.

__Known :__

Mass of block (m) = 1.5 kg

Acceleration due to gravity (g) = 10 m/s^{2}

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

Force (F) = 15 Newton

__Wanted:__ Net work is done on the object

__Solution :__

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) :

W_{1} = F s = (15 N)(2 m) = 30 N m = 30 Joule

Work done by horizontal component of weight (w_{x}) :

W_{2} = w_{x} s = (w sin 30^{o})(2 m) = (15 N)(1/2)(2 m) = 15 N m = 15 Joule

Net work done on the object :

W_{net} = W_{1} – W_{2} = 30 J – 15 J = 15 Joule

**What is the work-energy theorem in terms of net work and kinetic energy?****Answer**: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. Mathematically: $W_{net}=ΔKE$.

**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?****Answer**: 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.

**What is the gravitational potential energy of an object at ground level?****Answer**: 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.

**How does the kinetic energy of an object change if its velocity is doubled?****Answer**: 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.

**When is the gravitational potential energy of a falling object maximum?****Answer**: 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).

**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?****Answer**: 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.

**What happens to the gravitational potential energy of an object if it is raised to twice its original height above the ground?****Answer**: Gravitational potential energy is proportional to height. If an object is raised to twice its original height, its gravitational potential energy will double.

**When is the net work done on an object zero?****Answer**: 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.

**If an object is moving at a constant velocity, what can be inferred about the net work done on it over a certain distance?****Answer**: 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.

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