Physics 11 Dynamics and Momentum Sample Test answer key
1) Given the weight of an object is 843 N, how much net force is required on a frictionless surface to cause the object to accelerate at 4.25 m/s
2
? (2 marks)
Fnet = 843/9.8 * 4.25 = 366 N
2) A 47.9 N force is applied to a 13.2 kg block that only experiences an acceleration of 2.12 m/s
2
What is the coefficient of friction for the block on this horizontal surface? (3 marks)
Fnet = ma = 13.2 * 2.12 = 27.984 N
Fnet = Fapp + Ffriction
27.984 = 47.9 + Ffriction
Ffriction = -19.916 N coeff.ofFriction(MU)= Ffriction/Fnormal = 19.916 / (13.2*9.8) = 0.154
3) a)What applied force must be exerted to start accelerating a 62 kg block horizontally, if the coefficient of static friction is 0.65 ? (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 = 0.65 * 62 * 9.8 = 394.94 or 390 N b) If the kinetic coefficient is 0.44, what applied force is needed to accelerate the 62 kg block at a rate of 1.85 m/s 2 ? (3 marks)
Fnet = ma = 62 * 1.85 = 114.7 N
Ffriction = 0.44 * 62 * 9.8 = 267.344 N
Fnet = Fapp + Ffriction
114.7 = Fapp + -267.344
Fapp = 114.7 + 267.344 = 380N
4) A 4.2 kg fish is on one end of a line over a frictionless pulley, and a 6.5 kg tackle box is on the other end. What is the acceleration of the box if both are released ? (2 marks)
Fg1 = 4.2*9.8 = 41.16N Fg2 = 6.5 * 9.8 = 63.7 N
Fnet = 63.7-41.16 = 22.54 N a= 22.54 / (4.2.+6.5) = 2.106…m/s 2
What is the tension force in the line? (2 marks)
Looking at the box, Fg = 63.7 N down, a = 2.106…m/s 2 down, so Fnet = 6.5*2.106..= 13.6…N down
and Fnet = Fg + Tension so 13.6…= 63.7 + Tension and Tension = 13.6…-63.7 = - 50 N (up)
Looking at the fish, Fg = 41.16 N down, a = -2.106…m/s 2 up, so Fnet = 4.2*2.106..= -8.82 N up
and Fnet = Fg + Tension so -8.82…= 41.16 + Tension and Tension = -8.82 – 41.16 = - 50 N (up)
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5) What is the maximum acceleration possible lifting a 12.1 kg mass with a string that will break at 150
N of force ? (3 marks)
Fnet = Fg + T = 12.1 * 9.8 - 150 (use the maximum tension as the tension applied, then solving for )
Fnet = 118.58 - 150 = -31.42 N (the acceleration will give the maximum acceleration before breaking)
12.1 * a = -31.42 a = -2.6 m/s 2 where the negative upwards (since gravity was positive for Fg )
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5) A spring is compressed by 0.24 m when 42 kg is attached to it. What is k, the spring constant for this spring? (2 marks)
Fspring = -kx k = Fspring / x = (42 * 9.8 ) / 0.24 = 1715 N/m or 1700 N/m (or 17 N/cm)
How far will the spring compress if another 27 kg is added ? (2 marks) x = Fspring / k
= (27 + 42) * 9.8 / 1715 = 0.39 m
6) What is the force of gravity pulling on a 1900 kg satellite 188 million meters from earth's centre? (2 marks)
Fg = Gm
1 m
2
/(d 2 ) = 6.67 e -11 * 1900 * 5.97 e 24 / (188 e 6 ) 2 = 21 N
(estimate – on earth, 1900kg weighs 19000 N, 188 is about 30 time radius, so 900 times weaker)
7) What is the distance between the centers of a pair of 18.0 kg spheres if the force of gravity between them is 6.2 x10
-8
N ? (2 marks)
_6.2 x10 -8 = 6.67 e -11 *18 *18 / d 2 so d 2 = 6.67 e -11 *18 *18 / 6.2 x10 -8
d= sqrt(.348561) = 0.59 m
8) A 12 kg cart going 2.2 m/s collides with a 4.1 kg cart at rest. If the 4.1 kg cart moves off with a velocity of 6.5 m/s forward, what is the new velocity of the 12 kg cart? (2 marks)
12 * 2.2 + 4.1 * 0 = 12 v + 4.1 * 6.5
v = (12 * 2.2 - 4.1 * 6.5 ) / 12 v = -0.02 m/s (recoils backwards very slowly)
10) Floating out in space, a 111 kg astronaut (including suit mass) is holding on to a 375 kg piece of space junk, and both are drifting away from the shuttle at 01.70 m/s. The astronaut pushes hard against the junk and ends up moving towards the shuttle at 2.60 m/s. How fast is that piece of the space station now moving, and in what direction? (2 marks)
111 * 1.70 + 375 * 1.70 = 111 * -2.60 + 375 * v astronaut changes direction, v=negative
v = (111 * 1.70 + 375 * 1.70 - (111 * -2.60) ) / 375 v = 2.97 m/s (final answer is postitive, same direction as original, going faster since astronaut pushed off from it in other direction
11) A jet pack applies an average force of 2450 N for 3.22 seconds to a 152 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 5.00 seconds? (2 marks) a)Ft = m(v f
-v i
)
2450 * 3.22 = 152 (v - 0)
V=2450*3.22/152 = 51.9 m/s b) F*5.00 = 152(0-51.9…) be sure to use the unrounded version, (or use 2450*3.22 )
F = 1577.8 N or about 1580 N