Physics Engineering 1 (Phys. 213) Practice Exam 2 Solution, Feb. 22nd, 2008 (O'Shea)
Name __________________________________ Instructions: Please Read! This is a closed book, closed notes exam. This exam contains multiple-choice questions and problems. For each multiple-choice question, circle the letter of the best response. For problems, you need to show your work in a professional manner and write the final numerical answer with the proper unit in the space shown. Numerical answers without supporting calculations will receive zero credit. Partial credit is available only if your work is clear. 1. Academic honesty statement. This exam will NOT be marked unless you sign the academic honesty statement below. Your signature indicates that you have read, understood, and complied with the meaning of this statement. On my honor as a student I have neither given nor received unauthorized aid on this assignment.
Sign your name:__________________________________________________________
2. (1 point) Circle your Studio instructor/hour:
Lynch /TU 9:30
Matloob /TU 11:30
Regehr /TU 1:30
O'Shea/WF 7:30
Lynch /WF 9:30
Page total: /1
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3. A mass m of 2 kg is presses against a wall by a horizontal force of magnitude F as shown. The smallest value F can have without the mass slipping down the wall is 30 N. What is the coefficient of static friction between the mass and the wall? A freebody diagram is shown below. r y: Fs ,max - mg = 0 F m x: F - FN = 0 The friction force is Fs ,max = s FN = s F r Now using the y equation: Fs ,max s F - mg = 0 s = mg / F = 0.65 r r y FN F m
r mg
x
4. A worker pulls on a 12 kg crate with a force of magnitude 110 N at = 300 above the horizontal as shown. The coefficient of static friction between the crate and the floor is 0.60 and the coefficient of kinetic friction between the crate and the floor is 0.30.
F1
a) What is the value of fs,max, the maximum value of the static friction force? FN +F1sin mg = may = 0. f
FN
mg
F1
fS,max = sFN = s(mg -F1sin ) = (0.60) [(12 kg)(9.8 m/s2)- (110 N)sin300] = 37.6 N
y b) Determine the acceleration of the crate Check if the crate is stationary: find value of f required to keep crate stationary F1cos - f = 0 f = F1cos = 95.3 N
Since f > fs, max it does move. This means the crate moves and we are dealing with kinetic friction. fk = kFN= s(mg -F1sin ) Now summing forces along x (ax is not zero) F1cos -fk = max gives F1cos - k (mg -F1sin) = max So ax = [F1 cos - k (mg -F1sin)]/m = [110cos300 - 0.30 (12 x 9.8 -110sin300)]/12 = 6.4 m/s2 c) What is the value of the frictional force exerted on the floor by the crate (give direction as well) fk = 0.30 (12 x 9.8 -110sin300 )= 18.8 N, direction: right
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x
5. Two blocks are connected via a string over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. The angle is 30. Block A slides down the incline at constant speed. What is the mass of block B? Free body diagrams are shown below.
Using Newton's 2nd law sum forces on mB along y:
T - mB g = 0 T = mB g
[v = constant, a = 0) Now Using Newton's 2nd law we sum forces along x for mA: T + f k - m A g sin = 0 (1) [v = constant, a = 0) Summing forces along y:
fk mA
y x
y
mB
FN - m A g cos = 0
Since A is moving down the incline, the force of friction is uphill with magnitude
f k = k FN = k m A g cos
Substituting for fk and T in in equation (1)
m B = m A (sin - k cos ) = 3.27 kg
6. One end of a 1.0 m long string is fixed, the other end is attached to a 2.0 kg stone. The stone swings in a vertical circle, passing the bottom point at a speed of 4.0 m/s. a) What is the tension force exerted by the string on the stone at this point? Summing forces along y:
m B g + k m A g cos - m A g sin = 0
T - mg = ma y = mv 2 /r
T = m (g + v 2 /r ) = 2(9.8 + 4 2 / 1) = 51.6 N
y
r
T
mg
b) How much work does the tension force do on the stone during half a revolution? WT = 0 since the displacement is perpendicular to T
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7. You drop a 2.00 kg book to a friend who stands on the ground at distance D = 10.0 m below. If your friend's outstretched hands are at distance d = 1.50 m above the ground as shown (a) How much work does the gravitational force do on the book as it drops to her hands? Vertical displacement is 10.0 1.5 8.5 = m, downward.
v v Wg = Fg .d = mgd cos = ( 2.0)(9.8)(8.5) cos 00 = 167 J
(b) What is the change U in the gravitational potential energy of the bookEarth system during the drop? U = mgy = ( 2.0)(9.8)(- 8.5) = -167 J or you can use U = -Wg
(c) What is the speed of the book when it reaches her hands?
1 K + U = 0 0 - mv 2 + (- 167 J ) = 0 2 v = 2 167 / 2 = 12.9 m/s
8. A cord is used to raise a block of mass m with a constant upward acceleration g/4 through a vertical distance of 2.5 m Assume the block is initially stationary. (give answers below in terms of m, g, d). a) How much work is done by gravity?
v v Wg = Fg .d = mgd cos = - mgd
T
b) How much work is done by the tension force? T - mg = ma y = mg / 4
a
m
T = 5mg / 4 v v WT = T .d = 5mgd / 4
mg
y
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9. A 250 g block that is 1.2 m above the top of a relaxed spring is dropped onto the spring as shown. a) If the spring is compressed 0.5 cm before the block momentarily comes to rest what is the spring constant, k, of the spring. (h = 1.2 m, x = 0.005 m, m = 0.25 kg)
Work done by gravity Wg = mg(h + x) Work done by spring Ws = - kx2 Work-energy theorem K = W
(K f - K i ) = Wg + Ws
(0 - 0) = mg ( x + h) + - 1 kx 2 2 k = 2mg ( x + h) / x 2
(
)
2
Initial state vi=0 m
Final state
y
= 2 (0.25 kg ) 9.8 m/s 2 (0.005 + 1.2m ) / (0.005) = 236 kN/m
(
)
v1 h
m m
x
vf = 0
b) When the block momentarily comes to rest, what force does the spring exert on the block? F = -kx = (236 kN/m)(0.005 m ) = 1.18 kN
Multiple choice (circle or enter correct answer) v 10. A horizontal force F1 of magnitude 10 N is applied to a box on a floor, but the box does not v v slide. A second force F2 (points upward) is now applied and increased from zero. As F2 is increased do the following quantities increase, decrease, or stay the same (circle correct answer) v (a) magnitude of frictional force f s exerted on the box. Increase Decrease Stay same v (b) magnitude of normal force FN exerted on the box. Increase Decrease Stay same
(c) maximum value fs,max of the magnitude of the static frictional force exerted on the box? Increase (d) Does the box eventually slide?
Yes Decrease
Stay same
No
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y
11. A ball is whirled around in a circle on the end of a string. Its path is shown i) Give the directions of each of the following for the ball at position A (options are +x, -x, +y, -y)
C
x
velocity _-x__
acceleration _+y__
force__+y___ B
ii) Give the directions of each of the following for the ball at position B. velocity __+y_ acceleration _+x_ force___+x__
iii) When the ball is at position C, the string breaks. What direction does the ball move in just after the string breaks? ___+x___
A
12. A car with rear wheel drive (engine power goes to rear wheels only) accelerates to the east. During this acceleration the road applies a force to the rear wheels Frear and the road applies a force to the front wheels Ffront . Their directions are
a) Frear is east, Ffront is east c) Frear is west, Ffront is east
b) Frear is east, Ffront is west d) Frear is west, Ffront is west
13. An automobile drives in a circle of radius r0 at speed v0. The road is level and horizontal. Both r0 and v0 are now doubled. By what factor does the static friction force change?
a) increases by a factor of 4 d) decreases by a factor of 4
b) increases by a factor of 2 e) decreases by a factor of 2
c) does not change
14. Camping equipment on a sled (total weight 6000 N) is pulled across a frozen lake by means of a horizontal rope. The coefficient of kinetic friction between the sled and ice is 0.05. The work done by the campers in pulling the equipment 1000 m at constant velocity is:
a) 3.1 104 J
b) 1.5 105 J
c) 3.0 105 J
d) 2.9 106 J e) 6.0 106 J
f) none are correct
15. A block is released from rest at point P and slides along the frictionless track shown. At point Q, its speed is:
a) 2 g
d)
(h
1
- h2 )
b) 2 g (h1 - h2 )
c) (h1 - h2 ) / 2 g f) None are correct
2 g (h1 - h2 ) e) (h1 - h2 )2 / 2 g
16. A 700 N man jumps out of a window into a fire net 10 m below. The net stretches 2 m before bringing the man to rest and tossing him back into the air. The maximum potential energy of the net, compared to its unstretched potential energy, is:
a) 300 J
b) 710 J
c) 850 J
d) 7000 J 19
e) 8400 J
f) None are correct