**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

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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

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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

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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

[irp]