College Physics 4.4 – Measuring the mass of astronauts in orbit


Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s²
(a) Calculate her mass.
(b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which the recoil of the vehicle is avoided.


Solution:

Part A

From Newton’s second law of motion, we know that

\text{net Force}=\text{mass}\times \text{acceleration}

For this particular problem, we are solving for the mass. Hence, the formula for mass is 

\displaystyle \text{mass}=\frac{\text{net force}}{\text{acceleration}}

In symbols, that is 

\displaystyle \text{m}=\frac{\sum \text{F}}{\text{a}}

Substituting the given values and solving for the unknown, we have

\displaystyle \text{m}=\frac{50.0\:\text{N}}{0.893\:\text{m/s}^2}

\text{m}=55.9910\:\text{kg}

The mass of the austronaut is about 55.9910 kilograms. 

Part B

If the force could be exerted on the astronaut by another source other than the spaceship, then the spaceship would not experience a recoil. 


College Physics 4.3 – The acceleration of a laundry cart pushes by a cleaner


A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.


Solution:

From Newton’s second law of motion, we know that

\text{net Force}=\text{mass}\times \text{acceleration}

For this particular problem, we are solving for the acceleration. Hence, the formula for acceleration is 

\displaystyle \text{acceleration}=\frac{\text{net force}}{\text{mass}}

In symbols, that is 

\displaystyle \text{a}=\frac{\sum \text{F}}{\text{m}}

Substituting the given values and solving for the unknown, we have

\displaystyle \text{a}=\frac{60.0\:\text{N}}{4.50\:\text{kg}}

\text{a}=13.33\:\text{m/s}^2

The acceleration of the laundry cart is about 13.33 m/s2.


College Physics 2.64 – Consistency on the position and velocity diagrams


(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s.

(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64
Figure 2.65

Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.


College Physics 2.63 – Constructing a displacement graph


Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

2.63


Solution:

The position vs time graph is shown in the figure below. 

2.63b

College Physics 2.62 – Acceleration as the slope of the velocity vs time diagram


By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s²  at t = 10 s.

2.62
Figure 2.63

Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(0,\:165\right)\:and\:\left(20,\:228\right)

The velocityis computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}

a=3.2\:m/s^2

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.


College Physics 2.61 – Velocity as slope of position vs time


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}

v=0.238\:km/s

Therefore, the velocity is 0.238 km/s.


College Physics 2.60 – Slope of the position vs time diagram


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 10.0 s is 0.208 km/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}

v=0.208\:km/s

Therefore, the velocity is 0.208 km/s.


College Physics 2.59 – Analysis of motion diagrams


(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t = 20 s.

(b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s².

Figure 2.60
Figure 2.61

Solution:

Part A

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points \left(15,\:988\right)\:and\:\left(25,\:2138\right).

The slope is computed using the formula:

m=\frac{\Delta y}{\Delta x}

m=\frac{2138\:m-988\:m}{25\:s-15\:s}

m=115\:m/s

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds. 

Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(10,\:65\right)\:and\:\left(25,\:140\right)

The acceleration is computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}

a=5\:m/s^2

Therefore, the acceleration is verified to be 5.0 m/s².


Skateboarder on a ramp| Physics

A skateboarder starts up a 1.0-m-high, 30° ramp at a speed of 6.9 m/s. The skateboard wheels roll without friction. At the top, she leaves the ramp and sails through the air.

A) How far from the end of the ramp does the skateboarder touch down?

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Angular Acceleration| Circular Motion| Physics

Your car tire is rotating at 4.0 rev/s when suddenly you press down hard on the accelerator. After traveling 300 m, the tire’s rotation has increased to 6.5 rev/s . The radius of the tire is 32 cm.

A) What was the tire’s angular acceleration? Give your answer in rad/s²?

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