Category Archives: Physics

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 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

Grantham PHY220 Week 2 Problem 2

An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y axis and the total distance to the plane from you. Choose the x-axis east, y axis north, and z axis up.

Solution:

The angle of the plane in the x-y axis

Solve for \theta

\theta =tan^{-1}\left(\frac{1500}{2100}\right)=35.5^{\circ}

The angle is 35.5 degrees measured from the west axis. If it is measured from the East axis, the angle would be 180-35.5=144.5 degrees.

The distance from the plane to the observer

Solve for d=\sqrt{\left(1500\:m\right)^2+\left(2100\:m\right)^2+\left(5000\:m\right)^2}=5626.72\:m

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 2 Assignment

Problems:

  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]

  2. An airplane is 5,000 m above an observer and 2.1 km to the west of them and 1.5 km to the north of you. Determine the angle to the plane in the x – y-axis and the total distance to the plane from you. Choose the x-axis east, y-axis north, and z-axis up. [Solution]

  3. A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s? [Solution]

  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]

  5. If a 1500 kg car stopped from an in 5.6 seconds with an applied force of 5000 N, how fast was it initially traveling?[Solution]

  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]

  7. A 7.93 kg box is pulled along a horizontal surface by a force F_p of 84.0 N applied at a 47.0° angle. If the coefficient of kinetic friction is 0.35, what is the acceleration of the box? [Solution]

  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]

 

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.