Fisikastudycenter.com-Example Problem and Discussion on Enterprise and Energy, Materials Physics 2 class in high school. Include business relationships, style and movement, seek business from the difference in kinetic energy, determine the business of the difference in potential energy, negative positive sign on the business, total effort, and reading graphs F - S.
No Problem. 1
A retractable beam force F = 120 N that makes an angle of 37 ° to the horizontal direction.
If the beam shift as far as 10 m, determine the work done on the block!
Discussion
No Problem. 2
Beam mass 2 kg is on slippery surfaces accelerated from rest to move with an acceleration of 2 m / s 2.
Determine the work done on the beam during the 5 second!
Discussion
First look at the beam velocity 5 seconds, then look for the difference in the kinetic energy of the conditions of the crew and finally:
No Problem. 3
10 kg object about to be shifted through the slippery surface of the inclined plane as shown below!
Determine the effort required to move the object!
Discussion
Looking for a business with potential energy difference:
No Problem. 4
Note the chart style (F) against displacement (S) below!
Determine the size of the business up to 12 seconds!
Discussion
= The area between lines of business graph with axis S FS, for the graph above a trapezoidal area
W = 1/2 (12 + 9) x 6
W = 1/2 (21) (6)
W = 63 joules
(Thanks tuk Rora http://r-kubik-tu-rora.blogspot.com/ above correction)
No Problem. 5
A car of mass 5000 kg is moving with a speed of 72 km / h approaching a red light.
Determine the magnitude of the braking force to do so as the car stopped at a red light when it is 100 meters away from the car! (72 km / h = 20 m / s)
Discussion
No Problem. 6
A rod of length 40 cm and above-ground erect hammer sentenced to 10 kg from a height of 50 cm above the tip. When the average resistance force land 10 3 N, then the number of collisions hammer needs to be done in order to stick to the surface of the ground is flat ....
A. 4 times
B. 5 times
C. 6 times
D. 8 times
E. 10 times
(Problem UMPTN 1998)
Discussion
Two formulas effort involved here are:
In the hammer:
W = mg Δ h
On the ground by the force of friction:
W = FS
Find the depth of the entry of the stick (S) by one hammer blow:
FS = mgΔh
(10 3) S = 10 (10) (0.5)
S = 50/1000 = 5/100 m = 5 cm
So once the fall of the hammer, stick into the soil as deep as 5 cm. To 40 cm long stick, then the amount of the fall of the hammer:
n = 40: 5 = 8 times
No Problem. 7
A block is on an incline with a friction coefficient of 0.1 as shown in the following figure.
Beam drops down to 5 meters review.
Define:
a) the forces acting on the block
b) the business of each force on the beam
c) the total effort
Use g = 10 m / s 2, sin 53 o = 0.8, cos 53 o = 0.6, W (uppercase) for the symbol, and w (small) for the symbol gravity.
Discussion
a) the forces acting on the block
normal force (N), gravity (w) with its components, namely w sin w cos 53 ° and 53 °, the friction force F ges
b) the business of each force on the beam
With the incline as the trajectory (reference) displacement:
Normal-Businesses by force and gravity component w cos 53 °
The second effort is zero force (force perpendicular to the track)
-Business by gravity component w sin 53 °
W = w sin 53 °. S
W = mg sin 53 °. S
W = (6) (10) (0.8) (5) = + 240 joules
(Given a positive sign, mg sin 53 ° towards the direction of the displacement of the beam.)
-Effort by friction
Find great friction in advance
ges f = μ N
ges f = μ mg cos 53 °
ges f = (0.1) (6) (10) (0.6) = 0.36 N 3.6 N
W = - fges S = - 3.6 (5) = - 18 joules
(Given the negative sign, the direction opposite to the direction of the frictional force displacement of the beam)
c) the total effort
W total = +240 Joules - 18 joules = + 222 joules
Thanks for your submission to gita, .. the placement of a comma has been changed.
No Problem. 8
A block of mass 2 kg is on a rough inclined plane as shown in the following figure.
Beam pushed upward by a force F = 25 N to shift up to review the extent of 5 meters. Friction that occurs between the beams with the incline of 3 N. Slope of 53 ° to the horizontal plane.
Decide with positive or negative signs:
a) attempt by force F
b) attempt by friction
c) attempt to gravity
d) total effort
Discussion
a) attempt by force F
W = F. S = + 25 (5) = + 125 joules
b) attempt by friction
W = - f. S = - 3 (5) = - 15 joules
c) attempt to gravity
W = - mg sin 53 °. S = - (2) (10) (0.8) (5) = - 80 joules
d) total effort
W total = + 125 - 15-80 = 30 joules
Problem # 9
Object weighing 10 N is on a slippery incline with a slope angle of 30 °. When objects slide as far as 1 m, then the work done is the gravity ....
A. 10 sin 30 ° joule
B. 10 cos 30 ° joule
C. 10 sin 60 ° joule
D. 10 tan 30 ° joule
E. 10 tan 60 ° joule
(From about Ebtanas 1990)
Discussion
Effort by gravity
W = mg sin θ
From the problem has been known that (mg) = 10 Newton θ = 30 ° and, thus
W = 10 sin 30 ° joule
Problem # 10
A mass of 2 kg object falls freely from the top of a high rise building 100 m. If the friction with the air and the overlooked g = 10 ms -2 then the business that is done by gravity to a height of 20 m from the ground is .....
A. 200 joules
B. 400 joules
C. 600 joules
D. 1,600 joules
E. 2,400 joules
(From about Ebtanas 1992)
Discussion
Operations, changes in gravitational potential energy:
W = mgΔ h
W = 2 x 10 x (100-20)
W = 1600 joules
Problem # 11
A car with a mass of 1 ton moving from rest. A moment later the speed of 5 ms -1. Great work done by the engine is ...
A. 1,000 joules
B. 2,500 joules
C. 5,000 joules
D. 12,500 joules
E. 25,000 joules
(From Ebtanas 1994)
Discussion
Effort. change in kinetic energy of objects:
W = 1/2 m Δ (v 2)
W = 1/2 x 1 000 x 5 2
W = 12 500 joules
Note:
If you know two or velocity v, then v squared his first deducted, not deductible continues squared!.
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