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Dynamics Test v1
PSI Physics
Name____________________________________
Free Response - Solve each problem using one or more of these equations. Show all work.
∑F=ma
fk = μk FN
g = 9.8 m/s2
x = xo + vot + ½ at2
v = vo + at
v2 = vo2 + 2a (x - xo)
Use this information for Questions 1-7.
m1
m2
A 5.0 kg block, m1, lies on a horizontal tabletop. The coefficient of kinetic friction between the
block and the surface is 0.20. That block is connected by a massless string to a second block,
m2, whose mass is 3.5 kg. The string passes over a light frictionless pulley as shown above.
The system is released from rest.
1. Draw a clearly labeled free body diagram for m1. Include all forces and next to the free
body diagram draw the direction of the block’s expected acceleration, a1.
2. Draw a clearly labeled free body diagram for m2. Include all forces and next to the free
body diagram draw the direction of the block’s expected acceleration, a2.
3. Use Newton’s Second Law to write an equation for the tension (T) in the string based on
your free body diagram for m1.
4. Use Newton’s Second Law to write an equation for the tension (T) in the string based on
your free body diagram for m2.
5. Find the acceleration (a) of the system by simultaneously setting up and solving the
system of two equations.
6. Find the magnitude of the tension force in the string.
7. How fast will block m1 be moving 1.5 seconds after it is released?
---------------------------------------------------------------------------------------------------------------------8. A car is driven from rest to a speed of 25 m/s in 5 seconds. If the car has a mass of
1,200 kg, How much force did the engines exert on the car?
9. If a mass of 20 kg was hanging from a string, what would the Tension force be?
10. A person stands on a scale that reads 500 N. If the gravitational force at 500 N is the
action force, what and how much is the reaction force?
Dynamics Problem Test
Academic Physics
ANSWER KEY
convert to percent and apply 0.75 curve
Free Response 34 points
1.
+5 points
2.
+3 points
3. ΣF = m a
T – fk = m 1 a
T = fk + m 1 a
+3 points
4. ΣF = m a
T – m2 g = – m2 a
T = m2 g – m2 a
+3 points
5. T = fk + m1 a and T = m2 g – m2 a
+6 points
fk + m 1 a = m 2 g – m 2 a
μk m 1 g + m 1 a = m 2 g – m 2 a
a = (m2 g - μk m1 g) / (m1 + m2)
a = (3.5 kg 9.8 m/s2 – 0.2 (5 kg) 9.8 m/s2) / (5 kg + 3.5 kg)
a = 2.9 m/s2
6. T = m2 g – m2 a
T = 3.5 kg 9.8 m/s2 – 3.5 kg 2.9 m/s2
T = 24.15 N
7. v = vo + a t
v=at
v = 2.9 m/s2 1.5 s
v = 4.35 m/s
8. a =(vf – vi) / t
a = (25m/s – 0m/s) / 5s
a = 5m/s2
F = ma
F = 1200kg (5m/s2)
F = 6,000 N
9. T = mg
T = 20 kg (9.8m/s2)
T = 196 N
10. Reaction force = Normal Force = 500 N
+3 points
+3 points
+5 points
+3 points
+3 points