Download L:\FALL 2012\AIC 211 F12\211 EXAMS\Exam 2

PHYSICS 211 EXAM 2 RETAKE - ANSWERS
A___ 1.
World-class sprinters can accelerate out of the starting blocks with an acceleration that is
nearly horizontal and has magnitude 16 m/s2. How much horizontal force F must a sprinter
whose weight is 650 N exert on the starting blocks to produce this acceleration?
A. 1,060 N
C. 40.6 N
B. 10,400 N
D. 399 N
C___ 2.
If a car is moving to the left with constant velocity, one can conclude that
A. there must be no forces applied to the car.
B. the net force applied to the car is directed to the left.
C. the net force applied to the car is zero.
D. there is exactly one force applied to the car.
A___ 3.
Two forces are exerted on an object of mass m in the x direction as illustrated in the free-body
diagram. Assume that these are the only forces acting on the object. Which of the curves
labeled A to D on the graph could be a plot of vx(t), the velocity of the object in the x
direction as a function of time?
A. A
B. B
A___ 4.
C. C
D. D
Block B in the figure rests on a surface for which the static and kinetic coefficients of friction
are 0.59 and 0.42, respectively. The ropes are massless. What is the maximum mass of block
A for which the system is in equilibrium?
A. 14.75 kg
B. 10.50 kg
C. 25.00 kg
D. 42.37 kg
Problems: (20 points each) Show all work using correct notation and complete equations for full
credit.
5.
A box of mass 45.0 kg is being pulled up a rough incline by an applied force so that it accelerates
at 2.00 m/s2. The incline is tilted at 25.0O to the horizontal and the applies force is tilted 30.0O to
the incline surface. The coefficient of kinetic friction between the box and the surface is 0.35.
A. Draw a complete force diagram of all forces acting on the box.
B. List the forces on the box and express them in component form.
C. What is the magnitude of the applied force.
=
N
W
FA
fK
0iˆ
+
Njˆ
= −mg sin θ iˆ − mg cos θ ˆj
= FA cos φ iˆ + FA sin φ ˆj
=
− µ K Niˆ
+
0 ˆj
∑F
∑F
= N − mg cos θ + FA sin φ = 0 ⇒ N = mg cos θ − FA sin φ
Y
= − mg sin θ − FA cos φ − µ K N = ma ⇒ (substitute expression for normal)
−mg sin θ + FA cos φ − µ K ( mg cos θ − FA sin φ ) = ma ⇒ (solve for FA )
X
FA =
ma + mg sin θ + µ K mg cos θ
cos φ + µ K sin φ
FA = 399.9 N
⇒ (substitute values)
6.
A conical pendulum is formed by attaching a 0.250 kg ball to a 1.00 m-long string, then allowing
the mass to move in a horizontal circle of radius 48.0 cm. The figure shows that the string traces
out the surface of a cone, hence the name.
A. Draw a complete force diagram of all forces acting on the mass. Include force labels.
B. What is the speed of the mass?
C. What is the tension in the string?
T
W
= T sin θ iˆ + T cos θ ˆj
=
−
mg ˆj
0iˆ
sin θ =
R
L
R
 .48m 
O
⇒ θ = sin −1   θ = sin −1 
 = 28.69
L
 1m 
mv 2
∑ FX = T sin θ = R
∑ FY = T cos θ − mg = 0 ⇒ T cos θ = mg
divide equations:
v2
tan θ =
Rg
v = Rg tan θ
T=
mg
cos θ
solve for v
substitue values
substitute values
v = 1.6 ms
T = 2.79 N