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/s2. 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/s2
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/s2)(10 m) = 500 kg m2/s2 = 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 (vo) = 2 m/s
Final velocity (vt) = 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 = KE2 – KE1 = 1/2 m vt2 – 1/2 m vo2 = 1/2 m (vt2 – vo2) = 1/2 (2)(52-22) = (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/s2. What is the work done on the object.
Known :
Mass (m) = 10 kg
Initial velocity (vo) = 0 (initially at rest)
Time interval (t) = 3 seconds
Acceleration (a) = 2 m/s2
Wanted : Work (W)
Solution :
First, determine distance using the equation of the constant acceleration motion.
d = vo t + ½ a t2 = (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/s2 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/s2
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 (wx).
Work done by force (F) :
W1 = F s = (15 N)(2 m) = 30 N m = 30 Joule
Work done by horizontal component of weight (wx) :
W2 = wx s = (w sin 30o)(2 m) = (15 N)(1/2)(2 m) = 15 N m = 15 Joule
Net work done on the object :
Wnet = W1 – W2 = 30 J – 15 J = 15 Joule
[irp]
