Download PHYS 1P21/1P91 Quiz 4 Solutions 28 May 2013

PHYS 1P21/1P91
Quiz 4 Solutions
28 May 2013
1. A 2000-kg car is parked on a 20◦ slope. Determine the magnitude of the frictional force acting on
the car. [3 points]
Solution: First draw a free-body diagram (the forces are the weight mg, the normal force n, and
the frictional force f ) and select a coördinate system adapted to the ramp. Write Newton’s second
law for each direction, to obtain
n − mg cos 20◦ = 0
f − mg sin 20◦ = 0
The first equation is not needed for our purposes, as we only need to solve for f . Using the second
equation,
f = mg sin 20◦
f = (2000)(9.80)(0.342)
f = 6700 N
The magnitude of the frictional force acting on the car is 6700 N.
2. A 2000-kg car travelling at an initial speed of 50 km/h skids to a stop with a constant acceleration.
If the coefficient of kinetic friction between the locked tires and the road is µk = 0.50, determine the
length of the skid marks. [3 points]
Solution: The strategy is to use Newton’s second law to determine the acceleration of the car, and
then to use kinematics equations to determine the distance travelled by the car as it slows down and
stops.
From a free-body diagram of the car, we can conclude that the normal force acting on the car is
equal to its weight:
n = mg = 2000g
Thus, the frictional force acting on the car is
f = µk n = µk mg
Applying Newton’s second law to the horizontal motion of the car, we obtain
f = ma
µk mg = ma
a = µk g
a = 0.5g
a = 4.90 m/s
Using kinematics, we can determine the distance that the car travels while stopping:
(vx )2f = (vx )2i + 2ax ∆x
2
km
0 = 50
+ 2(−4.90)∆x
h
2
km 1000 m
1h
0 = 50
×
×
− 9.80∆x
h
1 km
3600 s
2
50 m
9.80∆x =
3.6 s
2
50
1
∆x =
m2 /s2
2 ×
3.6
9.80 m/s
∆x = 19.7 m
The skid marks are 19.7 m long.
3. Determine the acceleration of the 10-kg block if the coefficient of kinetic friction between the block
and the ramp is µk = 0.1. [4 points]
Solution: Draw a free body diagram for each of the blocks, and assume that the pulley is massless
and frictionless and that the strings are massless. For the 10-kg block, choose down the ramp as the
positive direction, and for the 5.0-kg block choose upwards as the positive direction. By adopting
this convention, the accelerations of the two blocks are identical (including the sign), so we can use
the same symbol a for each of them.
From the free-body diagram for the 10-kg block, we obtain the equations:
n = m1 g cos 40◦
m1 g sin 40◦ − T − f = m1 a
(2)
(1)
Because the coefficient of kinetic friction is 0.1, it follows that
f = µk n = (0.1)m1 g cos 40◦
(3)
The free-body diagram for the 5.0-kg block yields the equation
T − m2 g = m2 a
(4)
Substituting the expressions for f and T from Equations (3) and (4) into equation (2), we obtain
m1 g sin 40◦ − (m2 g + m2 a) − ((0.1)m1 g cos 40◦ ) = m1 a
m1 g sin 40◦ − m2 g − m2 a − (0.1)m1 g cos 40◦ = m1 a
m1 g sin 40◦ − m2 g − (0.1)m1 g cos 40◦ = m1 a + m2 a
m1 g sin 40◦ − m2 g − (0.1)m1 g cos 40◦
a=
m1 + m2
◦
10g sin 40 − 5g − g cos 40◦
a=
10 + 5
10 sin 40◦ − 5 − cos 40◦
a=
·g
15
a = 0.43 m/s
The magnitude of the acceleration is 0.43 m/s2 , and it is directed down the ramp.