Walker3 P. 045. [544707] Show Details

Дата канвертавання27.04.2016
Памер62 Kb.
Chapter 5-6
1. Walker3 5.P.045. [544707] Show Details In a tennis serve, a 0.070 kg ball can be accelerated from rest to 35 m/s over a distance of 0.70 m. Find the magnitude for the average force exerted by the racket on the ball during the serve.61.3 N


Newton’s second law: F = m a

To find F, we need to find a:

For constant acceleration motion: s = (vf2 – vi2 ) /(2*a)

Rearrange the equation: a = (vf2 – vi2 )/(2*s) = (352-02)/(2*0.70)=875m/s2

Therefore, the average force is: F= m a = 0.070*875=61.3N

2. Walker3 5.P.046. [544509] Show Details A 48.0 kg swimmer with an initial speed of 1.80 m/s decides to coast until she comes to rest. If she slows with constant acceleration and stops after coasting 2.00 m, what was the magnitude of the force exerted on her by the water?38.9 N


Similar to problem 1:

Acceleration a = (vf2-vi2)/(2*s) = (0-1.802)/(2*2.00) =-0.81m/s2

Force F = m a = 48.0 * (-0.81) = -38.9N

Therefore, the magnitude of the force is 38.9N

3. Walker3 5.P.048. [544489] Show Details An object of mass m = 6.10 kg has an acceleration a = (1.17 m/s2)x + (-0.664 m/s2)y. Three forces act on this object: F1, F2, and F3. Given that F1 = (3.50 N)x and F2 = (-1.55 N)x + (2.05 N)y, find F3.( 5.19 N)x + ( -6.1 N)y


Newton’s second law: ΣF = m a  F1 + F2 + F3 = m a

Therefore, F3 = m a – F1 – F2

F3= 6.10*(1.17 x-0.664 y) - 3.50 x{-1.55 x + 2.05 y}

= 5.19 x -6.1y (N)
4. Walker3 5.P.049. [544617] Show Details At the local grocery store, you push a 12.6 kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N, you now accelerate the cart from rest through a distance of 2.49 m in 3.00 s. What was the mass of the dog food? 9.09 kg


Newton’ second law: F = m a

To find m, we need to know F and a

For constant acceleration motion: xf = xi + vit +1/2at2=0+0+1/2a*3.002=2.49

Therefore, a = 2.49*2/3.002 = 0.553m/s2

The total mass is, m = F/a = 12.6 + mdog_food­ = 12.0/0.553

Thus, mdog_food= 9.09 kg

5. Walker3 5.P.050. [544540] Show Details Biomechanical research has shown that when a 85 kg person is running, the force exerted on each foot as it strikes the ground can be as great as 2315 N. (a) What is the ratio of the force exerted on the foot by the ground to the person's body weight? 2.78 (b) If the only forces acting on the person are (i) the force exerted by the ground and (ii) the person's weight, what is the magnitude and direction of the person's acceleration?


17.4 m/s2


---Select--- upward downwardupward

(c) If the acceleration found in part (b) acts for 11.5 ms, what is the resulting change in the vertical component of the person's velocity? 0.2 m/s


  1. W = m g = 85*9.81 N,

the ratio = 2315/(85*9.81) = 2.78

  1. ΣF = ma = Fground + W 

a =(Fground + W)/m = ( 2315 y + (-85*9.81) y)/85 = 17.4m/s2 y (upward)

(c) ΔV = a * Δt = 17.4 y * 11.5e-3 = 0.2 m/s

6. Walker3 5.P.051. [544654] Show Details To become airborne, a 2.0 g grasshopper requires a takeoff speed of 2.4 m/s. It acquires this speed by extending its hind legs through a distance of 4.3 cm. (a) What is the average acceleration of the grasshopper during takeoff? 67 m/s2 (b) Find the magnitude of the average net force exerted on the grasshopper by its hind legs during takeoff. 0.134 N (c) If the mass of the grasshopper decreases, does the takeoff acceleration increase, decrease, or stay the same? stays the same (d) If the mass of the grasshopper decreases, does the required takeoff force increase, decrease, or stay the same?decreases Explain.


  1. average acceleration a = (vf2-vi2)/(2*s)=2.42/(2*4.3e-2) = 67m/s2

  2. F = ma = 2.0e-3 * 67 = 0.134N

  3. Stay the same: a is not related to mass in (a)

  4. Decrease, since F = m a, m decreases  F decreases

7. Walker3 5.P.052. [544661] Show Details On an aircraft carrier, a jet can be catapulted from 0 to 145 mi/h in 2.00 s. If the average force exerted by the catapult is 5.60 106 N what is the mass of the jet?1.73e+05 kg

Average acceleration = (vf – vi)/ Δt = 145*1609/3600 / 2.00 = 32.4m/s2

F = ma  m = F/a = 5.60e6/32.4=1.73e5 kg

8. Walker3 5.P.054. [544618] Show Details An apple of mass m = 0.15 kg falls out of a tree from a height h = 3.3 m.(a) What is the magnitude of the force of gravity, mg, acting on the apple? 1.47 N (b) What is the apple's speed, v, just before it lands?8.05 m/s

(c) Calculate mgh and (mv2)/2, including both their numerical values and their units. Express their units in terms of kg, m, and s. 4.85595 kg m^2/s^2 [mgh] 4.85595 kg m^2/s^2 [(mv2)/2] (The relationship between these two quantities will be investigated in Chapter 8.)

  1. mg = 0.15 *9.81 = 1.47N

  2. h = vf2/2g  vf = sqrt(2*h*g) = sqrt(2*3.3*9.81) = 8.05m/s

  3. mgh = 0.15*9.81*3.3 = 4.86 kg m^2/s^2

1/2mv2 = 1/2*0.15*8.052 = 4.86kg m^2/s^2
9. Walker3 5.P.056. [544680] Show Details Paleontologists estimate that if a Tyrannosaurus rex were to trip and fall, it would have experienced a force of approximately 256,200 N acting on its torso when it hit the ground. Assume the torso has a mass of 3740 kg. (a) Find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about 7g.) 58.7 m/s2 (b) Assuming the torso is in free fall for a distance 1.59 m as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground? 0.0952 s

  1. ΣF = m a = F – mg  a = (F-mg)/m = 256200/3740-9.81 = 58.7m/s2

  2. the touch down velocity = sqrt ( 2*s*g) = sqrt (2.0*1.59*9.81)

The time for the torso to come to rest = touch down velocity / acceleration

= sqrt(2.0*1.59*9.81) / 58.7 = 0.0952 s

10. Walker3 5.P.022. [544536] Show Details Two crewmen pull a boat through a lock, as shown in Figure 5-25. One crewman pulls with a force of F1 = 130 N at an angle of = 37° relative to the forward direction of the raft. The second crewman, on the opposite side of the lock, pulls at an angle of 45°. With what force F2 should the second crewman pull so that the net force of the two crewmen is in the forward direction? 111 N

Figure 5-25

Assume right, up to be positive:

Total force in parallel = -F1 * sin 37° + F2 * sin 45°
Total force in parallel = 0, therefore; F2 = 111N

11. Walker3 5.P.044. [544714] Show Details An ant walks slowly away from the top of a bowling ball, as shown in Figure 5-31. If the ant starts to slip when the normal force on its feet drops below one-sixth its weight, at what angle does slipping begin?80.4°

Figure 5-31

Normal force = mg*cos = 1/6*mg  cos = 1/6  =80.4°

12. Walker3 6.P.053. [549557] Show Details A skateboard accident leaves your leg in a cast and supported by a traction device, as in Figure 6-27, with = 33.0°. Find the mass m that must be attached to the rope if the net force exerted by the small pulley on the foot is to have a magnitude of 32 N.1.94 kg

Figure 6-27

Magnitude of the force F =T1 + T2 = T1 *cos 33° + T2 *cos 33° = 32N

Tension force is the same on the string, T1 = T2

T2 = mg


32N = 2*mg*cos33°  m = 32/2/9.81/cos33°

m = 1.94kg
13. Walker3 6.P.055. [549553] Show Details Find the coefficient of kinetic friction between a 3.55 kg block and the horizontal surface on which it rests if a 86.0 N/m spring must be stretched by 5.50 cm to pull it with constant speed. Assume that the spring pulls in the horizontal direction.

Fr = Fspring

Fspring = k x = 86.0*5.5e-2 N

N = mg

Fr = N *μk μk= Fr/N = Fspring / mg = 86.0*5.5e-2/(3.55*9.81) = 0.136
14. Walker3 6.P.057. [549574] Show Details When a block is placed on top of a vertical spring, the spring compresses 3.90 cm. Find the mass of the block, given that the force constant of the spring is 1800 N/m. 7.16 kg
The forces acting on the block are: gravity (downward) and spring elastic force (upward). The total force = 0 = Fspring – mg = k x – mg,


m = kx / g = 1800 * 3.90e-2 / 9.81 = 7.16 kg
15. Walker3 6.P.060. [549580] Show Details A force of 13.5 N pulls horizontally on a 1.3 kg block that slides on a rough, horizontal surface. This block is connected by a horizontal string to a second block of mass m2 = 1.63 kg on the same surface. The coefficient of kinetic friction is µk = 0.27 for both blocks.(a) What is the acceleration of the blocks? 1.96 m/s2 (b) What is the tension in the string? 7.51 N

To estimate the friction force, we need to know the normal force.

Fr = N * k

for the first block, N1 = m1 g = 1.3 * 9.81 N = 12.8 N

for the second block, N2 = m2 g = 1.63 * 9.81 N = 16.0 N

Fr1 = N1 * 0.27 = 3.44N; Fr2 = N2 * 0.27 = 4.31N

the net force on the two block system is

F – Fr1 – Fr2 = 13.5N – 3.44N – 4.31N = 5.75 N

the acceleration is

a = F / m = 5.75 /(m1+m2) = 5.75/(1.3+1.63) = 1.96 m/s2


considered the second block,

the total force acting on it = T– Fr2 = m2 a = 1.96*1.63 = 3.20N

therefore, the tension force T = 4.31 + 3.20 N = 7.51 N

16. Walker3 6.P.062. [549516] Show Details A 10 g coin slides upward on a surface that is inclined at an angle of 14° above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.23; the coefficient of static friction is 0.31. Find the magnitude and direction of the force of friction under the following circumstances. (a) when the coin is sliding 0.0219 N down the incline (b) after it comes to rest
0.0237 N up the incline

(a) kinetic friction Fr = N * k = mg*cos(14°) * k = 0.0219N

direction is against motion, downward;

(b) static friction force Fr = N * s = mg*cos(14°) * s = 0.0237N

direction is against the tendency of motion, upward

17. Walker3 6.P.065. [549542] Show Details A ball of mass m is placed in a wedge, as shown in Figure 6-37, in which the two walls meet at a right angle. Assume the walls of the wedge are frictionless.

Figure 6-37

(a) Determine the magnitude of contact force 1. (Use g for acceleration due to gravity and m as necessary.) F1 = mg sin(45-theta) (b) Determine the magnitude of contact force 2. (Use g for acceleration due to gravity and m as necessary.) F2 = mg cos(45-theta)



The two contact forces are perpendicular to each other. The angle between the first contact force and negation of the gravity is 45° + , thus

contact force 1 = mg * cos(45° + ) = mg sin(45° - )


Angle betwen force 2 and negation of the gravity is 45° - , therefore

contact force 2 = mg * cos(45° - )
18. Walker3 6.P.066. [549531] Show Details A 2.0 kg box rests on a plank that is inclined at an angle of 65° above the horizontal. The upper end of the box is attached to a spring with a force constant of 16 N/m, as shown in Figure 6-38. If the coefficient of static friction between the box and the plank is 0.18, what is the maximum amount the spring can be stretched and the box remain at rest?1.2 m

Figure 6-38
At the maximum stretch, the spring force is at its maximum. Since gravity and normal force are not changing. Since it is in translation equilibrium, all forces add up to zero. Therefore, the sum of friction force and spring force is constant. Therefore the friction force must be at its maximum on the opposite direction to the spring force.

Maximum friction force = N * s = mg cos(65°) * 0.18 = 1.49N

Along the incline plane, all forces projections add up to zero


Fspring – Fr – mg sin(65°) = 0


Fspring = k x


x = Fspring / k = (Fr + mgsin(65°))/ k = 1.2 m

19. Walker3 6.P.068. [549565] Show Details The blocks shown in Figure 6-39 are at rest.

Figure 6-39

Given that the mass of block A is 8.80 kg and the coefficient of static friction between block A and the surface on which it rests is 0.300, what is the maximum mass block B can have and the system still remain in equilibrium? 2.64 kg

translational equilibrium:

considering the junction:

There tension forces add up to zero.

The horizontal tension force = vertical tension force = mB g

considering block A

The horizontal tension = Friction force = N * s = mA g * s = mB g


mB = mA * s = 8.80 * 0.300 = 2.64 kg

20. Walker3 6.P.075. [549566] Show Details In a daring rescue by helicopter, two men with a combined mass of 174 kg are lifted to safety. (a) If the helicopter lifts the men straight up with constant acceleration, is the tension in the rescue cable greater than, less than, or equal to the combined weight of the men? greater (b) Determine the tension in the cable if the men are lifted with a constant acceleration of 1.13 m/s2? 1.9 kN
(a) greater, before the net force is upward which means the tension force is greater than the downward gravity force.

(b) Newton's second law,

ΣF = m a = T – G = T – mg


T = ma + mg = m (a+g) = 174 *(1.13+9.81) = 1.9 e3 N = 1.9 kN
21. Walker3 6.P.074. [549538] Show Details Find the coefficient of kinetic friction between a 4.2 kg block and the horizontal surface on which it rests if an 70 N/m spring must be stretched by 1.7 cm to pull the block with constant speed. Assume that the spring pulls in a direction 11° above the horizontal.0.0285
To find out the k, we need to know normal force and kinetic friction force.
on the vertical direction, all forces projections sum up to zero

N – G + Fspring * sin 11° = 0

Fspring = k x

so the normal force N = 4.2 * 9.81 - 70*0.017* sin 11°

the the horizontal direction, the net force

Fspring 11° - Fr

so the friction force Fr = 70*0.017* cos 11°
The friction coefficient is

k = Fr / N = 0.0285

22. Walker3 6.P.079. [549544] Show Details Find the acceleration that the cart in Figure 6-44 must have in order for the cereal box at the front of the cart not to fall. Assume that the coefficient of static friction between the cart and the box is 0.33. 29.7 m/s2

Figure 6-44


Here, the normal force = m a

Friction force = Gravity = mg = ma * s


a = g / s = 9.81/0.33 = 29.7 m/s2

23. Walker3 6.P.083. [549564] Show Details As your plane circles an airport, it moves in a horizontal circle of radius 2250 m with a speed of 355 km/h. If the lift of the airplane's wings is perpendicular to the wings, at what angle should the plane be banked so that it doesn't tend to slip sideways?23.8°
the centripetal force = mg * tan

the centripetal acceleration = g * tan = v2/r


tan = v2/rg


= arctan ( v2/rg ) = 23.8 °

24. Walker3 6.P.088. [549540] Show Details Find the tension in each of the two strings shown in Figure 6-32 for general values of the masses. (Use m1 for the mass m1, m2 for m2, m3 for m3, and g for g as necessary.) left string (m1 m3 g) / (m1+m2+m3)
right string ((m1+m2) m3 g) / (m1+m2+m3)

Figure 6-32
similar to the solution for the 2 body system

a = m3 g / (m1 + m2 + m3)

left string, tension force = m1 a = (m1 m3 g) / (m1+m2+m3)

right string, tension force = (m1 + m2) a = ((m1+m2) m3 g) / (m1+m2+m3)

База данных защищена авторским правом ©shkola.of.by 2016
звярнуцца да адміністрацыі

    Галоўная старонка