PCS120 Review and Additional Practice for Midterm (with solutions)

Knight 3rd edition

1.18, 1.36, 1.38, 1.41, 2.1, 2.3, 2.20, 2.43, 2.47, 2.67, 5.28, 5.15, 6.19, 6.28, 6.31, 6.53, 6.54

1.18.

(a) Dot Time (s) x (m) (b)

1 0 0 2 2 30 3 4 95 4 6 215 5 8 400 6 10 510 7 12 600 8 14 670 9 16 720

1.36. Represent (Sam + car) as a particle for the motion diagram.

1.38. Represent the speed skater as a particle for the motion diagram.

1.41. Represent the car as a particle for the motion diagram.

2.1. Cars will be treated by the particle model.

Beth and Alan are moving at a constant speed, so we can calculate the time of arrival as follows:

1 0 1 01 0

1 0

x x x x xv t t

t t t v

Using the known values identified in the pictorial representation, we find:

Alan 1 Alan 0Alan 1 Alan 0

Beth 1 Beth 0Beth 1 Beth 0

400 mile8:00 AM 8:00 AM 8 hr 4:00 PM

50 miles/hour400 mile

9:00 AM 9:00 AM 6.67 hr 3:40 PM60 miles/hour

x xt t

vx x

t tv

(a) Beth arrives first. (b) Beth has to wait Alan 1 Beth 1 20 minutest t for Alan.

Times of the order of 7 or 8 hours are reasonable in the present problem.

2.3. (a) The time for each segment is 1 50 mi/40 mph 5/4 hrt and 2 50 mi/60 mph 5/6hr.t The average speed

to the house is 100 mi

48 mph5/6 h 5/4 h

(b) Julie drives the distance 1x in time 1t at 40 mph. She then drives the distance 2x in time 2t at 60 mph. She

spends the same amount of time at each speed, thus

1 2 1 2 1 2/40 mph /60 mph (2/3)t t x x x x

But 1 2 100 miles,x x so 2 2(2/3) 100 miles.x x This means 2 60 milesx and 1 40 miles.x Thus, the

times spent at each speed are 1 40 mi/40 mph 1.00 ht and 2 60 mi/60 mph 1.00 h.t The total time for her return

trip is 1 2 2.00 h.t t So, her average speed is 100 mi/2 h 50 mph.

2.20. We will use the particle model and the constant-acceleration kinematic equations.

(a) Substituting the known values into 211 0 0 2

,y y v t a t we get

2 21 1

110 m 0 m 20 (m/s) ( 9.8 m/s )

2t t

One of the roots of this equation is negative and is not relevant physically. The other root is 1 4 53 s,t which is the answer

to part (b). Using 1 0 ,v v a t we obtain 2

1 20(m/s) ( 9.8 m/s )(4.53 s) 24 m/sv

(b) The time is 4.5 s.

A time of 4.5 s is a reasonable value. The rock’s velocity as it hits the bottom of the hole has a negative sign because of its downward direction. The magnitude of 24 m/s compared to 20 m/s, when the rock was tossed up, is consistent with the fact that the rock travels an additional distance of 10 m into the hole.

2.43. The car is a particle and constant-acceleration kinematic equations hold.

(a) This is a two-part problem. During the reaction time, 2

1 0 0 1 0 0 1 0( ) 1/2 ( )

0 m (20 m/s)(0.50 s 0 s) 0 m 10 m

x x v t t a t t

After reacting, 2 1 110 m 10 m 100 m,x x that is, you are 100 m away from the intersection.

(b) To stop successfully, 2 2 2 2 22 1 1 2 1 1 12 ( ) (0 m/s) (20 m/s) 2 (100 m) 2 m/sv v a x x a a

(c) The time it takes to stop once the brakes are applied can be obtained as follows: 2

2 1 1 2 1 2 2( ) 0 m/s 20 m/s ( 2 m/s )( 0.50 s) 11 sv v a t t t t

The total time to stop since the light turned red is 11.5 s.

2.47. Model the flea as a particle. Both the initial acceleration phase and the free-fall phase have constant acceleration, so use the kinematic equations. :

We can apply the kinematic equation 2 2f i 2v v a y twice, once to find the take-off speed and then again to find the final

height. In the first phase the acceleration is up (positive) and 0 0.v 2 2 31 0 1 0 12 ( ) 2(1000 m/s )(0.50 10 m) 1.0 m/sv a y y v

In the free fall phase the acceleration is 1a g and 1 1 0 m/sv and 2 0 m/s.v 2 2 2 22 1 1

2 1 21

(1.0 m/s)5.1 cm

2 2( ) 2( 9.8 m/s )

v v vy y

a g

So the final height is 2 15 1 cm 5 1 cm 0 50 mm 5 2 cm.y y

This is pretty amazing–about 10–20 times the size of a typical flea.

2.67. Jill and the grocery cart will be treated as particles that move according to the constant-acceleration kinematic equations.

The final position of Jill when the cart is caught is given by

2 2 2 2J1 J0 J0 J1 J0 J0 J1 J0 J0 J1 J1

1 1 1( ) ( ) 0 m 0 m ( 0 s) (2.0 m/s )

2 2 2x x v t t a t t a t t

The cart’s position when it is caught is

2 2 2C1 C0 C0 C1 C0 C0 C1 C0 C1

2 2C1

1 1( ) ( ) 20 m 0 m (0.5 m/s )( 0 s)

2 2

20 m (0.25 m/s )

x x v t t a t t t

t

Since J1 C1x x and J1 C1,t t we get

2 2 2 2 2J1 C1 C1 C1

2 2 2 2C1 C1

1(2.0) 20 s 0.25 0.75 20 s 5.16 s

2

20 m (0.25 m/s ) 20 m (0.25 m/s )(5.16 s) 26.7 m

t t t t

x t

So, the cart has moved 6.7 m.

5.28.

Figure (a) shows velocity as downward, so the object is moving down. The length of the vector increases with each step showing that the speed is increasing (like a dropped ball). Thus, the acceleration is directed down, as shown. Since

F ma

the force is in the same direction as the acceleration and must be directed down.

Figure (b), however, shows the velocity as upward, so the object is moving upward. But the length of the vector decreases with each step showing that the speed is decreasing (like a ball thrown up). Thus, the acceleration is also directed down, as shown. As in part (a), the net force must be directed down

5.15.

Please refer to Figure EX5.15. Newton’s second law is .F ma The graph tells us the acceleration as a function of mass. Knowing the mass and acceleration for any given point, we can find the force. We chose the m = 600g = 0.60 kg, which gives a = 6.0 m/s2. Newton’s

second law yields 2(0.60 kg)(6.0 m/s ) 3.6 NF ma

Assess: To double-check the result insert F = 3.6 N into Newton’s law for m = 200 g = 0.20 kg. This gives 2(3.6 N)/(0.2 kg) 18 m/s ,a F m which is consistent with the graph.

6.19.

We will represent the crate as a particle.

(a) When the belt runs at constant speed, the crate has an acceleration 20 m/sa

and is in dynamic equilibrium. Thus

net 0.F

It is tempting to think that the belt exerts a friction force on the crate. But if it did, there would be a net force

because there are no other possible horizontal forces to balance a friction force. Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and the gravitational force on the crate. (A friction force would have been needed to get the crate moving initially, but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’s motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maximum possible value s max s .f n The maximum possible acceleration of the crate

is

s max smax

f na

m m

If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the free-body diagram that Gn F mg Thus,

6.28.

You can model the beam as a particle in static equilibrium.

Using Newton’s first law, the equilibrium equations in vector and component form are:

net 1 2 G

net 1 2 G

net 1 2 G

0 N

( ) 0 N

( ) 0 Nx x x x

y y y y

F T T F

F T T F

F T T F

Using the free-body diagram yields:

1 1 2 2 1 1 2 2 Gsin sin 0 N cos cos 0 NT T T T F

The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, 1T and 2T

Substituting 1 220 , 30 , and G 9800 N,F mg the simultaneous equations become

1 2 1 2sin 20 sin30 0 N cos20 cos30 9800 NT T T T

You can solve this system of equations by simple substitution. The result is 1 6397 N 6400 NT and 2 4376 NT

4380 N. The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope.

6.31.

We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic equations.

(a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and another acceleration as the elevator brakes from 10.0 s to 12.0 s. The weight is the same as the gravitational force during constant velocity motion, so Henry’s weight Gw F mg is 750 N. His weight is less than the gravitational force on him during

the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the elevator’s initial motion is down.

(b) Because the gravitational force on Henry is 750 N, his mass is G/ 76.5 kg 77 kg.m F g

(c) The apparent weight during vertical motion is given by

G

1 1a w

w mg a gg F

During the interval 0 s 2 s,t the elevator’s acceleration is

2600 N1 1.96 m/s

750 Na g

At 2 s,t Henry’s position is

2 21 0 0 0 0 0

1 1( ) ( ) 3.92 m

2 2y y v t a t a t

and his velocity is

1 0 0 0 3.92 m/sv v a t a t

During the interval 22 s 10 s, 0 m/s .t a This means Henry travels with a constant velocity 1 3.92 m/s.v At

10 st he is at position

2 1 1 1 35.3 my y v t

and he has a velocity 2 1 3.92 m/s.v v During the interval 10 s 12.0 s,t the elevator’s acceleration is

2900 N1 1.96 m/s

750 Na g

The upward acceleration vector slows the elevator and Henry feels heavier than normal. At 12 0 st Henry is at position

23 2 2 2 2

1( ) ( ) 39.2 m

2y y v t a t

Thus Henry has traveled distance 39.2 m 39 m.

6.53.

The antiques (mass )m in the back of your pickup (mass )M will be treated as a particle. The antiques touch the

truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques system will also be treated as a particle, and the contact force on this particle will be due to the road.

(a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass ,m M Newton’s second law is

net G PA( ) (( ) ) ( ) 0 N 0 N ( ) ( )x x x x x xF F N F f f m M a m M a

net G PA( ) (( ) ) ( ) N ( ) 0 N 0 Ny y y y yF F N F f m M g

The model of static friction is ,f N where is the coefficient of friction between the tires and the road. These

equations can be combined to yield .a g Since constant-acceleration kinematics gives 2 21 0 1 02 ( ),v v a x x we find

2 2 2 21 0 0

min 21 0 1 0

(25 m/s)0.58

2( ) 2 ( ) (2)(9.8 m/s )(55 m)

v v va

x x g x x

The truck cannot stop if is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80

respectively (see Table 6.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for min. This

value is smaller than the given coefficient of static friction s( 0.60) between the antiques and the truck bed. Therefore,

the antiques will not slide as the truck is stopped over a distance of 55 m. The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol Web site, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck.

6.54 The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the model of kinetic friction.

The normal force due to the wall, which is perpendicular to the wall, is here to the right. The box slides down the wall at

constant speed, so 0a

and the box is in dynamic equilibrium. Thus, net 0.F

Newton’s second law for this equilibrium situation is

net push( ) 0 N cos 45xF n F

net k push G k push( ) 0 N sin 45 sin 45yF f F F f F mg

The friction force is k k .f n Using the x-equation to get an expression for n, we see that k k push cos 45 .f F

Substituting this into the y-equation and using Table 6.1 to find k 0.20 gives,

k push pushcos 45 sin 45 0 NF F mg

2

pushk

(2.0 kg)(9.80 m/s )23 N

cos45 sin 45 0.20cos45 sin 45

mgF