Physics 11 Dynamics and Momentum Sample Test II ___
Name : ________________
30
Dynamics formulas:
Fnet = ma and Fnet = sum of all external forces
FNormal = m x 9.80 N/kg FFriction =μFNormal
-11
2
Fg =6.67 x 10 m1 m2 / d mass of Earth = 5.97 x 1024 kg radius of Earth= 6.38 x 106 m
1) Given the weight of an object is 348 N, how much net force is required on a frictionless surface to
cause the object to accelerate at 1.25 m/s2 ? (2 marks)
Fnet = 348/9.8 * 1.25 = 44.4 N
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2) A 57.9 N force is applied to a 11.2 kg block that only experiences an acceleration of 3.12 m/s2 What
is the coefficient of friction for the block on this horizontal surface? (3 marks)
Fnet = ma = 11.2 * 3.12 = 34.944 N
Fnet = Fapp + Ffriction
34.944 = 57.9 + Ffriction
Ffriction = -22.956 N
coeff.ofFriction(MU)= Ffriction/Fnormal = 22.956 / (11.2*9.8) = 0.209
3) a)What applied force must be exerted to start accelerating a 12 kg block horizontally, if the
coefficient of static friction is 0.45 ? (1 mark)
not yet moving means Fnet = 0 and Fnet = Fapp + Ffriction, so the force of friction IS the force
applied to get it to just start moving Ffriction = MU Fnormal = 0.45 * 12 * 9.8 = 52.92 or 53 N
b) If the kinetic coefficient is 0.34, what applied force is needed to accelerate the 12 kg block at a
rate of 2.85 m/s2 ? (3 marks)
Fnet = ma = 12 * 2.85 = 34.2 N
Ffriction = 0.34 * 12 * 9.8 = 39.984 N
Fnet = Fapp + Ffriction
34.2 = Fapp + - 39.984
Fapp = 34.2 + 39.984 = 74 N
4) A 3.3 kg fish is on one end of a line over a frictionless pulley, and a 6.7 kg tackle box is on the other
end. What is the acceleration of the box if both are released ? (2 marks)
Fg1 = 3.3*9.8 = 32.34 N Fg2 = 6.7 * 9.8 = 65.66 N
Fnet = 65.66 – 32.34 = 33.32 N
a= 33.32 / (3.3 +6.7) = 3.332 m/s2
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What is the tension force in the line? (2 marks)
Looking at the box, Fg = 65.66 N down, a = 3.332 m/s2 down, so Fnet = 6.7*3.332 = 22.3…N down
and Fnet = Fg + Tension so 22.3…= 65.66 + Tension and Tension = 22.3…-65.66 = - -43.3… N (up)
Looking at the fish, Fg = 32.34 N down, a = -3.332…m/s2 up, so Fnet = 3.3*3.332= - 10.9956 N up
and Fnet = Fg + Tension so - 10.9…= 32.34 + Tension and Tension = - 10.9… – 32.34 = - 43.3… N (up)
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5) What is the maximum acceleration possible lifting a 15 kg mass with a string that will break if
subjected to 190 N of force ? (3 marks)
Fnet = Fg + T = 15 * 9.8 - 190 (use the maximum tension as the tension applied, then solving for )
Fnet = 147 - 190 = -43 N (the acceleration will give the maximum acceleration before breaking)
15 * a = -43
a = -2.866… = -2.9 m/s2 where the negative upwards (since gravity was positive for Fg )
5) A spring is compressed by 2.5 cm when 0.52 kg is attached to it. What is k, the spring constant for
this spring? (2 marks)
Fspring = -kx k = Fspring / x = (0.52 * 9.8 ) / 2.5 = 2.0384 N/cm
6)
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If this spring is compressed by 3.7 cm, what force is pushing on it? (2 marks)
Fspring = k*x
= 2.0384 N/cm *.3.7 cm = 7.5 N or a mass of 0.7696 kg attached
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What is the force of gravity pulling on a 760 kg satellite 98 million meters from earth's centre? (2
marks)
Fg = Gm1m2/(d2) = 6.67 e -11 * 760 * 5.97 e 24 / (98 e 6 )2 = 31.51… = 32 N
(estimate – on earth, ~7600 N, 98e6 is about 15 time radius, so 225 times weaker)
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What is the distance between the centers of a pair of 7.0 kg spheres if the force of gravity between
them is 8.2 x10-8 N ? (2 marks)
8.2 x10-8 = 6.67 e -11 *7 *7 / d2 so d2 = 6.67 e -11 *7 *7 / 8.2 x10-8
d= sqrt(0.0398) = 0.1996… = 0.20 m
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A 6.0 kg cart going 2.2 m/s collides with a 2.1 kg cart at rest. If the 2.1 kg cart moves off with a
velocity of 4.5 m/s forward, what is the new velocity of the 6.0 kg cart? (2 marks)
6 * 2.2 + 2.1 * 0 = 6 v + 2.1 * 4.5
v = (6 * 2.2 - 2.1 * 4.5 ) / 6
v = 0.625 m/s (still forward)
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10) Floating out in space, a 121 kg astronaut (including suit mass) is holding on to a 175 kg piece of
space junk, and both are drifting away from the shuttle at 2.70 m/s. The astronaut pushes hard
against the junk and ends up moving towards the shuttle at 1.60 m/s. How fast is that piece of the
space junk now moving, and in what direction? (2 marks)
121 * 2.70 + 175 * 2.70 = 121 * -1.60 + 175 * v
astronaut changes direction, v=negative
v = (121 * 2.70 + 175 * 2.70 - (121 * -1.60) ) / 175
v = 5.67 m/s (final answer is postitive, same direction as original, going faster since astronaut
pushed off from it in other direction
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11) A jet pack applies an average force of 1350 N for 4.00 seconds to a 255 kg mass at rest. What is
the velocity of the mass immediately after the jet pack finishes? What force would now be needed to
reduce the momentum of the mass back to zero in 12.00 seconds? (2 marks)
a)Ft = m(vf-vi)
1350 * 4.0 = 255 (v - 0)
V=1350*4.00/255 = 21.17…m/s
b) F*12.00 = 255(0-21.17…) be sure to use the unrounded version,
F = 450 N or exactly 1/3 of 1350 since we use 3* the time!