The block cannot provide any frictional force to the left because it is resting on a frictionless table. Whether the block is allowed to move or not, the tension force in the string is all generated by the suspended \(\displaystyle \small 200g\) mass. The force in the string is equal and opposite to the gravitational force acting on the suspended mass.

\(\displaystyle F_T=-F_g=-mg\)

\(\displaystyle F_T=-(0.2kg)(-9.8\frac{m}{s^2})=1.96N\)

The system is in equilibrium when no parts are in motion. Alternatively stated, the upward tension in the string is equal and opposite to the downwards force on the mass. If the block is not allowed to move as more mass is added, at some point the tensile force in the string will exceed its mechanical limits, and the string will break.

For the problem we need to understand Newton's second law: \(\displaystyle F_n_e_t= ma\). The net upward forces and net downward forces must equal the product of mass and acceleration. Since the box is traveling upwards with an acceleration of 1m/s² we can indicate upward forces as positive and downward forces as negative. Indicating T as tension and Fg as the weight of the box, we can find the value of tension with the equation \(\displaystyle T -F_g=ma\).

\(\displaystyle T=ma+F_g=(20kg)(1\frac{m}{s^2})+(20kg)(10\frac{m}{s^2})=220N\)

Gravitational acceleration is related to distance via the equation:

\(\displaystyle a=\frac{GM}{r^2}\)

\(\displaystyle \small G\) is the gravitational constant, \(\displaystyle \small M\) is the mass of the attracting object, and \(\displaystyle \small r\) is the distance from its center.

In this case, the initial distance (on the surface) is \(\displaystyle \small 1R\) from the center of planet X, and the distance of the satellite is \(\displaystyle \small 4R\) from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.

\(\displaystyle a_{surface}=\frac{GM}{R^2}=15\frac{m}{s^2}\)

\(\displaystyle a_{satellite}=\frac{GM}{(4R)^2}=\frac{GM}{16R^2}\)

\(\displaystyle a_{satellite}=\frac{1}{16}\frac{GM}{R^2}\)

By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.

\(\displaystyle a_{satellite}=\frac{1}{16}(a_{surface})=\frac{1}{16}(15\frac{m}{s^2})\)

\(\displaystyle a_{satellite}=\frac{15}{16}\frac{m}{s^2}\)

To solve this question, we will need to use the equation for acceleration:

\(\displaystyle a=\frac{v_2-v_1}{t}\)

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.

\(\displaystyle v_2=-v_1\)

\(\displaystyle v_2-v_1=-2v_1=-2v\)

Plug this value into the equation for acceleration.

\(\displaystyle a=\frac{-2v}{t}\)

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.

\(\displaystyle a_E=g=\frac{-2v}{t}\rightarrow t_{E}=\frac{-2v}{g}\)

\(\displaystyle a_M=\frac{g}{6}=\frac{-2v}{t}\rightarrow t_M=6\frac{-2v}{g}\)

We can see that the time of flight on the moon is equal to six times the time of flight on Earth.

The force due to gravity on any object can be given by the equation below.

\(\displaystyle F = \frac{GMm}{r^{2}}\)

\(\displaystyle G\) is the gravitational constant, \(\displaystyle M\) is the mass of the earth, \(\displaystyle m\) is the mass of the object, and \(\displaystyle r\) is the distance between the center of each object.

In our question, the only value to change is the radius of the new planet; both masses and \(\displaystyle G\) remain constant. The effect of doubling the radius on the force is given below.

\(\displaystyle F_2=\frac{GMm}{(2)^2}=\frac{1}{4}F_1\)

The person's weight on the new planet would be one-fourth their weight on Earth.

\(\displaystyle \frac{1}{4}1000N=250N\)

To relate force and mass, we use Newton's second law:

\(\displaystyle F=ma\)

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.

\(\displaystyle F_m=ma_m\)

\(\displaystyle F_m=100N\ \text{and}\ a_m=\frac{1}{6}g\)

\(\displaystyle 100N=m(\frac{1}{6}g)\)

\(\displaystyle 100N=m(\frac{1}{6}*9.8\frac{m}{s^2})\)

\(\displaystyle m=\frac{100N}{1.63\frac{m}{s^2}}=61.22kg\)

Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.

It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, F_g, downwards, and the normal force of the scale, F_n, upwards.

Use these to write the net force equation.

\(\displaystyle \small F_{net} = ma = F_{n}-F_{g}\)

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that F_n = 500N, and we can solve for m.

\(\displaystyle \small 0=F_{n}-mg = 500-mg\).

\(\displaystyle \small m=\frac{500N}{10\frac{m}{s^{2}}}=50kg\)

When the elevator comes to rest, then we have an acceleration of -0.5 m/s² (downwards acceleration). Plugging this in, we can solve for the new F_n.

\(\displaystyle \small (50kg)(-0.5 \frac{m}{s^{2}})=F_{n}-(50kg)(10\frac{m}{s^{2}})\)

\(\displaystyle \small F_{n}=500-25 = 475 N\)

First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, F_N in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.

Assuming downward is the positive y direction, we can solve for F_N.

mg – F_N = ma

F_N = mg – ma = 0 N

Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity. 

The acceleration of gravity is given by the equation \(\displaystyle a_{g} = \frac{GM}{r^{2}}\), where G is constant.

For Earth, \(\displaystyle a_{g} = \frac{GM_{earth}}{r_{earth}^{2}} = g\).

For the new planet, 

\(\displaystyle a_{g} = \frac{G(9M_{earth})}{\left ( 3r_{earth} \right )^2} = \frac{G(9M_{earth})}{9r_{earth{}}^2} = \frac{GM_{earth}}{r_{earth}^{2}}} = g\).

So, the acceleration is the same in both cases.

The force between two objects is calculated by:

\(\displaystyle F_G=\frac{Gm_{1}m_{2}}{d^2}\)

Use our given variables:

\(\displaystyle F_G=\frac{G(4M)(M)}{D^2}=4\frac{GM^2}{D^2}\)