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HOW TO SOLVE PHYSICS PROBLEMS
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THE PROCEDURE Step 1:Draw a free body diagram Step 2:Write down the givens Step 3:Write down the unknown Step 4:Resolve the free body diagram
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THE PROCEDURE Step 5:Use equations for the x and y directions Step 6:Use “sum of” substitution Step 7:Plug in givens and solve for unknown Step 8:Check the answer with common sense
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Problem 11 (page 128) Two forces, F1 and F2, act on a 5.00 kg block. The magnitudes of the Forces are F1=45.0 N and F2=25.0 N. What is the horizontal acceleration of the block? F1F1 F2F2 65
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Problem 11 (page 128) § Step 1: Draw free body diagram. F2F2 65 F1F1 + y x
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Problem 11 (page 128) m = 5.00 kg F1 = 45.0 N F2 = 25.0 N §Step 2: Write down the givens.
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Problem 11 (page 128) Horizontal Acceleration = a x §Step 3: Write down the unknown.
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Problem 11 (page 128) §Step 4: Resolve the free body diagram. F2F2 65 F1F1 + y x F 2x F 2y
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Problem 11 (page 128) Step 5: Use equations for the x and y directions. F2 65 F1 + y x ∑ Fx=ma ∑ Fy=ma
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Problem 11 (page 128) F 1x - F 2 = m ax F 1y = m ay §Step 6: Use ”sum of” substitution.
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F 1 cos 65 -F 2 = a x m F 1 cos 65 -F 2 = a x m Problem 11 (page 128) §Step 7: Plug in givens and solve.
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Problem 11 (page 128) -1.20 m/s §Step 8: Check with common sense. 2 Sounds good!
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Problem 55 (page 131) A helicopter is traveling horizontally to the right at constant velocity. The Weight of the helicopter is 53800 N. The lift force L generated by the rotating blade makes an angle of 21 degrees with respect to the verticle. What is the magnitude of the lift force and determine the magnitude of the air resistance R that opposes the motion.
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Problem 55 (page 131) § Step 1: Draw free body diagram. L R W 21 + y x
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Problem 55 (page 131) W = 53800 N § Step 2: Write down the givens.
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Problem 55 (page 131) The lift force (L) Air resistance (R) § Step 3: Write down the unknown.
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Problem 55 (page 131) § Step 4: Resolve the free body diagram. L R W 21 + y xLxLx LyLy
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Problem 55 (page 131) Step 5: Use equations for the x and y directions. R W 21 + y xLxLx LyLy ∑ F x = ma ∑ F y = ma
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Problem 55 (page 131) L x - R = ma L y - W = ma § Step 6: Use ”sum of” substitution.
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W = L (cos 21) Problem 55 (page 131) § Step 7: Plug in givens and solve. L sin 21 = R
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Problem 55 (page 131) L = 57627 N R = 20651 N § Step 8: Check with common sense.
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Problem 79 (page 133 ) The weight of a block on a table is 111 N and that of the hanging block is 258 N. Assuming the coefficient between the 111 N block and the table is 0.300, find the acceleration of the blocks and the tension of the string.
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Problem 79 (page 133 ) § Step 1: Draw free body diagram. + N W1W1 F T T W2W2
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Problem 79 (page 133 ) u =0.300 W1 = 111 N W2 = 258 N § Step 2: Write down the givens.
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Problem 79 (page 133 ) Acceleration of the two blocks (a) Tension of the string (T) § Step 3: Write down the unknown.
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Problem 79 (page 133 ) § Step 4: Resolve the free body diagram. + N W1W1 F T T W2W2
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Problem 79 (page 133 ) Step 5: Use equations for the x and y directions. + N W1W1 F T T W2W2 ∑Fx = ma ∑ Fy = 0 ∑ Fx = ma
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Problem 79 (page 133 ) T-F = ma W 1 -N = 0 W 2 -T = ma § Step 6: Use ”sum of” substitution.
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T = m 1 (a + ug) a = W 2 -T m 2 Problem 79 (page 133 ) § Step 7: Plug in givens and solve.
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Problem 79 (page 133 ) T = 100.89 N a = 5.974 m/s § Step 8: Check with common sense. 2
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Problem 111 (page 135) A 45 kg box is sliding up an incline that makes an angle of 15 degrees with respect to the horizontal. The coefficient of friction between the box and the surface of the incline is 0.180. The initial speed of the box at the bottom of The initial speed of the box at the bottom of the incline is 1.5 m/s, How far does the box travel along the incline before coming to rest.
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Problem 111 (page 135) § Step 1: Draw free body diagram. 15 N F W +
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Problem 111 (page 135) m = 45 kg u = 0.180 V o = 1.5 m/s § Step 2: Write down the givens.
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Problem 111 (page 135) Distance the box travels (d) § Step 3: Write down the unknown.
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Problem 111 (page 135) § Step 4: Resolve the free body diagram. N F W WxWx WyWy 15
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Problem 111 (page 135) Step 5: Use equations for the x and y directions. N F W WxWx WyWy 15 ∑ Fx = ma ∑ Fy = 0
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Problem 111 (page 135) - F - W x = ma N - W y = 0 § Step 6: Use ”sum of” substitution.
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a = -4.245 m/s d = V - V o 2a Problem 111 (page 135) § Step 7: Plug in givens and solve. 2 22
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Problem 111 (page 135) § Step 8: Check with common sense. d = 0.265 m
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by ANDREW MEDLEY
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