Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43

Grantham PHY220 Week 2 Assignment Problem 7

A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47^{\circ}  angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box?

SOLUTION:

The free-body diagram of the box

week 2 problem 7

Solve for the normal force

\sum F_y=ma_y

F_N-mg+F_psin\left(47.0^{\circ} \right)=0

F_N-7.93\left(9.80\right)+84.0\:sin\left(47^{\circ} \right)=0

F_N=16.28\:N

Solve for the friction force

F_{fr}=\mu _kF_N=0.35\left(16.28\:N\right)=5.70\:N

Solve for the acceleration in the horizontal direction

\sum F_x=ma_x

F_p\:cos\:\left(47^{\circ} \right)-F_{fr}=7.93\left(a_x\right)

84.0\:N\cdot cos\:\left(47^{\circ} \right)-5.70\:N=7.93\:kg\:\cdot \left(a_x\right)

a_x=\frac{51.59\:N}{7.93\:kg}=6.51\:m/s^2

Therefore, the acceleration of the box is 6.51 m/s2 along the horizontal surface.

Grantham PHY220 Week 2 Assignment Problem 6

If the acceleration due to gravity on the Moon is 1/6 that what is on the Earth, what would a 100 kg man weight on the Moon? If a person tried to simulate this gravity in an elevator, how fast would it have to accelerate and in which direction?

SOLUTION:

The acceleration due to gravity on the moon is

g_m=\frac{1}{6}\left(9.80\:m/s^2\right)=1.63\:m/s^2

The weight of a 100-kg man on the moon is

W_m=mg_m=\left(100\:kg\right)\left(1.63\:m/s^2\right)=163.3\:N

If the elevator is accelerating upward then the acceleration would be greater. The person would be pushed toward the floor of the elevator making the weight increase. Therefore, the elevator must be going down to decrease the acceleration.

For a 100 kg man to experience a 163.3 N in an elevator,

F=ma

163.3\:N=100\:kg\:\left(9.80\:m/s^2-a_e\right)

9.80-a_e=\frac{163.3}{100}

a_e=9.80-\frac{163.3}{100}

a_e=8.167\:m/s^2

Therefore, the elevator should be accelerated at 8.167 m/s2 downward for a 100-kg man to simulate his weight just like his weight in the moon which has 1/6 of the Earth’s gravity acceleration.

Grantham PHY220 Week 2 Assignment Problem 4

A not so brilliant physics student wants to jump from a 3rd-floor apartment window to the swimming pool below. The problem is the base of the apartment is 8.00 meters from the pool’s edge. If the window is 20.0 meters high, how fast does the student have to be running horizontally to make it to the pool’s edge?

Solution:

Since the student will be running horizontally, there is no initial vertical velocity, v_{0_y}=0. We are also given \Delta x=8\:m, and \Delta y=-20\:m.

Consider the vertical component of the motion.

\Delta y=v_{0_y}t-\frac{1}{2}gt^2

-20=0-\frac{1}{2}\left(9.80\right)t^2

-20=-4.9t^2

t^2=\frac{20}{4.9}

t=\sqrt{\frac{20}{4.9}}

t=2.02\:s

Consider the horizontal component of the motion

\Delta x=v_{0_x}t

v_{0_x}=\frac{\Delta x}{t}

v_{0_x}=\frac{8}{2.02}

v_{0_x}=3.96\:m/s

Therefore, the student should be running 3.96 m/s horizontally to make it to the pool’s edge.

Grantham PHY220 Week 2 Assignment Problem 1

A ship has a top speed of 3 m/s in calm water. The current of the ocean tends to push the boat at 2 m/s on a bearing of due South. What will be the net velocity of the ship if the captain points his ship on a bearing of 55° North of West and applies full power?

Solution:

Week 2 Problem 1

R_x=-3\:cos\:55^{\circ }=-1.720729309\:m/s

R_y=3\:sin\:55^{\circ }-2=0.4574561329\:\:m/s

The x component of the resultant is negative and the y component is positive, thus the resultant is located at the second quadrant.

R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2}=\sqrt{\left(-1.720729309\:m/s\right)^2+\left(0.4574561329\right)^2}=1.78\:m/s

\theta =tan^{-1}\left(\frac{R_y}{R_x}\right)=tan^{-1}\left(\frac{0.4574461329}{-1.720129309}\right)=-14.9^{\circ}

Therefore, the magnitude of the net velocity of the ship is 1.78 m/s, and is going 14.9 degrees North of West

Grantham PHY220 Week 1 Assignment Problem 8

Problem 8

A stone is dropped from the roof of a high building. A second stone is dropped 1.25 s later. How long does it take for the stones to be 25.0 meters apart?

Solution:

Let t be the amount of time after the first stone is dropped. The distance from traveled by the first stone is

y_1=\frac{1}{2}gt^2

The distance traveled by the second stone is

y_2=\frac{1}{2}g\left(t-1.25\right)^2

The difference between the two stones is 25.0 m after time

y_1-y_2=25.0

\frac{1}{2}\left(9.80\:m/s^2\right)t^2-\frac{1}{2}\left(9.80\:m/s^2\right)\left(t-1.25\right)^2=25

4.9t^2-4.9\left(t^2-2.5t+1.5625\right)=25.0

4.9t^2-4.9t^2+12.25t-7.65625=25.0

12.25t=25.0+7.65625

12.25t=32.65625

t=2.67\:s

 

Grantham PHY220 Week 1 Assignment Problem 7

Problem 7

Explain a possible situation where you start with a positive velocity that decreases to a negative increasing velocity while there is a constant negative acceleration.

Solution:

An example of this situation is a free-fall. If an object is thrown upward, the initial velocity is positive. Then the velocity decreases until the object thrown will reach its maximum height and then it goes back with a negative increasing velocity. In this entire flight, the acceleration is a constant negative–the acceleration due to the Earth’s gravity.

 

Grantham PHY220 Week 1 Assignment Problem 5

Problem 5

A car travels 120 meters in one direction in 20 seconds. Then the car returns ¾ of the way back in 10 seconds. Calculate the average speed of the car for the first part of the trip. Find the average velocity of the car.

Solution:

The average speed of the car for the first part of the trip is

average\:speed=\frac{x}{t}=\frac{120\:m}{20\:s}=6\:m/s\:

The average velocity of the car is 

\overline{v}=\frac{\Delta x}{\Delta t}=\frac{120-\frac{3}{4}\left(120\right)\:m}{20+10\:s}=\frac{30\:m}{30\:s}=1\:m/s