Homework 4 (323506)

  

 

 

  About this Assignment

 

Due: Fri Oct 7 2005 11:32 PM EDT

Description
Chapter 6

 

 

pts

1

-

  1. Walker2 6.P.001. [236035] A baseball player dives into third base with an initial speed of 8.40 m/s. If the coefficient of kinetic friction between the player and the ground is 0.39, how far does the player slide before coming to rest?

Work done by non-conservative force = change in total mechanical energy

In this problem, there is no change in potential energy (Player slides on level ground).

Change in total mechanical energy = change in kinetic energy

F x = K(final) – K(initial) = 0 – (1/2)Mv2, where v = initial velocity

Frictional force opposes motion, if x>0, then F<0

F = - m N

In vertical direction, acceleration is zero.  Therefore net vertical force is zero.

N – Mg = 0

N = Mg

F =- m Mg

Work = Fx = - m Mg x = - K(initial) =  – (1/2)Mv2,

 m g x =   (1/2)v2,

Notice that the mass M and the minus sign cancel from both sides.

We don’t need to know the mass of the baseball player, given our model of friction.

x = v2 / [ 2 m g ] = (8.40 m/s)2 / [ 2* 0.39 * 9.80 m/s2] = 9.23 m
[9.23] m

 

 

pts

1

-

 

 

  1. Walker2 6.P.007. [236038] To move a large crate across a rough floor, you push down on it at an angle of 21°, as shown in Figure 6-16. Find the force necessary to start the crate moving, given that the mass of the crate is m = 30 kg and the coefficient of static friction between the crate and the floor is µ S= 0.52.

FNet = m a
FNet,y = ma y = 0
N-Mg - Fsin(21°) = 0
N = Mg+Fsin(21°)

FNet,x = max = 0
Fcos(21°)-FS = 0
FS at maximum, just before slipping
FS =µS N
Fcos(21°) -µS ( Mg  + F sin(21°) ) = 0
F [ cos(21°) -µS sin(21°) ] = µS  Mg
F = µS   Mg / [ cos(21°) -µS sin (21°) ]
F =  0.52 ( 30 kg) (9.81 m/s 2) / [ 0.9336-(0.52)(0.3584) ]
F = 204.8 kg m/s2 = 205 N

[205] N
Figure 6-16

 

 

 

pts

1

-

  1. Walker2 6.P.015. [236040] Pulling down on a rope, you lift a 4.06 kg bucket of water from a well with an acceleration of 2.05 m/s2. What is the tension in the rope?

Net force on Bucket:

T – Mg = Ma

T = M(a+g)  = (4.06 kg) (2.05 m/s2 + 9.80 m/s2) = 48.1 kg m/s2
[48.1] N

 

 

pts

1

-

4. Walker2 6.P.018. [236042] A 53.3 kg person takes a nap in a (lightweight) backyard hammock. Both ropes supporting the hammock are at an angle of 12.6° above the horizontal. Find the tension in the ropes.
2 T sin( 12.6°) - Mg = M a
a = 0 (unless ropes break a sleeping person falls to ground).
2 T sin( 12.6 ° ) - Mg = 0
T = Mg / [ 2 sin( 12.6 ° ) ] =  (53.3 kg) (9.81 m/s2) / [ 2 (0.218)] = 1198 kg m/s 2
[1200] N

 

 

pts

1

-

2

-

3

-

5. Walker2 6.P.019. [236043] A backpack full of books weighing 52.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 165 N/m is attached to the backpack and pulled horizontally, as indicated in Figure 6-17.


Figure 6-17

(a) If the spring is pulled until it stretches 2.30 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table?
F-fs = 0
F = - kx
k= 165 N/m
x = -2.30 cm = -0.0230 m
fs = F = - kx = - (165 N/m) ( -0.0230 m) = 3.79 N
[3.79] N
(b) Does your answer to part (a) change if the mass of the backpack is tripled?

yes  

no  
Explain.

Key: The frictional force was determined using only the spring constant and the extension of the spring.

 

 

pts

1

-

2

-

3

-

6. Walker2 6.P.021. [236045] The pulley system in Figure 6-18 is used to lift a crate of mass m = 51 kg. Note that a chain connects the upper pulley to the ceiling and a second chain connects the lower pulley to the crate. Assume that the masses of the chains, pulleys, and ropes are negligible.


Figure 6-18

(a)   Determine the force F required to lift the crate with constant speed.

Tension in rope = F

Net force on Crate =  F+F-Mg,  notice that the rope pulls twice on the crate.

Acceleration of crate = 0

2F-Mg = 0

F = Mg/2 = (51 kg) (9.80m/s2) / 2 = 250 N
[250] N
(b) Determine the tension in each chain when the crate is being lifted with constant speed.

Top chain:  T-2F = 0, T=500N

Bottom chain: T-Mg = 0,  T=500 N
[500] N (top chain)
[500] N (bottom chain)

 

 

 

pts

1

-

2

-

7. Walker2 6.P.030. [236049] In Figure 6-21 we see two blocks connected by a string and tied to a wall, with = 35°. The mass of the lower block is m = 0.8 kg; the mass of the upper block is 2.0 kg.


Figure 6-21

(a) Find the tension in the string connecting the two blocks.
FNet = m a
FNet,y = ma y = 0
N+mg sin(270 ° -35°) = 0
N = - mgsin(270-35°)  (Not actually needed for solution)

F Net,x = max = 0
T + mg cos(270° -35°) =
T = - m g cos(270-35°) =  - (0.8 kg) (9.81 m/s2) cos(235 °) = + 4.50 N

[4.5] N
(b) Find the tension in the string that is tied to the wall.
Repeat identical analysis, but replace mass with (0.80 kg + 2.0 kg) = 2.80 kg
T = - (0.8 kg + 2.0 kg)  g cos(270-35°) =  +15.8 N
[15.8] N

 

 

 

 

pts

1

-

8. Walker2 6.P.032. [236051] Two blocks are connected by a string, as shown in Figure 6-23. The smooth inclined surface makes an angle of = 42° with the horizontal, and the block on the incline has a mass m1 = 7.1 kg. Find the mass m2 of the hanging block that will cause the system to be in equilibrium.

m2:  T- = ma = 0

m1:  x-axis points up the incline, y-axis is perpendicular to incline

The force of gravity makes an angle q1 = -90°-q counterclockwize with respect to the x-axis

q1<0 means the angle is clockwize.

x-component of net force on m1:

Fnet,x = m1 g cos(-90°-q ) + T

Acceleration = 0

Fnet,x = 0

m1 g cos(-90°-q ) + T = 0

m1 g cos(-90°-q ) + m2g =0

m2 = - m1cos(-90°-42° )= -(7.1 kg)[cos(-132deg)] = -(7.1 kg)(-0.669) =+4.75 kg


[4.75] kg


Figure 6-23

 

 

Submit for Testing