Forces

Help Questions

Physics › Forces

Questions 1 - 10

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are apart, what is the gravitational force between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.99*10^{11}N$

$3.01*10^{-8}N$

$5.98*10^{12}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are apart, what is the gravitational force between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.99*10^{11}N$

$3.01*10^{-8}N$

$5.98*10^{12}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

Two asteroids in space are in close proximity to each other. Each has a mass of $6.69*10^{15}kg$ . If they are apart, what is the gravitational force between them?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.99*10^{11}N$

$3.01*10^{-8}N$

$5.98*10^{12}N$

$1.49*10^{-32}N$

$1.11*10^{30}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=(6.67*10^{-11}\frac{m^3}{kg\cdot s^2})(\frac{(6.69*10^{15}kg )(6.69*10^{15}kg )}{(100,000m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(\frac{4.48*10^{31}kg^2}{10*10^9m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}(4.48*10^{21}\frac{kg^2}{m^2})$

$F_G=2.99*10^{11}N$

An astronaut lands on a planet with twelve times the mass of Earth and the same radius. What will be the acceleration due to gravity on this planet, in terms of the acceleration due to gravity on Earth?

$\frac{1}{144}g$

$\frac{1}{12}g$

$g\sqrt{12}$

Explanation

For this comparison, we can use the law of universal gravitation and Newton's second law:

$F=G\frac{m_1m_2}{r^2}$

We know that the force due to gravity on Earth is equal to . We can use this to set the two force equations equal to one another.

$G\frac{m*m_{E}}{r^2}=mg$

Notice that the mass cancels out from both sides.

$G\frac{m_{E}}{r^2}=g$

This equation sets up the value of acceleration due to gravity on Earth.

The new planet has a mass equal to twelve times that of Earth. That means it has a mass of . It has the same radius as Earth, . Using these variables, we can set up an equation for the acceleration due to gravity on the new planet.

$G\frac{12m_{E}}{r^2}=a$

$12(G\frac{m_{E}}{r^2})=a$

We had previously solved for the gravity on Earth:

$G\frac{m_{E}}{r^2}=g$

We can substitute this into the new acceleration equation:

The acceleration due to gravity on this new planet will be twelve times what it would be on Earth.

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass , are $10*10^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.53*10^{-5}N$

$4.74*10^{-6}N$

$3.55*10^{-6}N$

$4.61*10^{-10}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

It actually doesn't matter which asteroid we're looking at; the gravitational force will be the same. This makes sense because Newton's 3rd law states that the force one asteroid exerts on the other is equal in magnitude, but opposite in direction, to the force the other asteroid exerts on it.

A car rounds a perfectly circular turn at a constant speed. This causes the acceleration to .

remain constant

increase

decrease

become zero

not be predictable

Explanation

Acceleration results from a change in velocity. Despite the speed remaining constant, velocity is a vector quantity and will change if the car changes direction. In rounding the turn, there is a change in the direction of the velocity, but not in the magnitude. This change in direction causes a non-zero acceleration.

The acceleration will remain equal to the equation for centripetal acceleration:

$a_c=\frac{v^2}{r}$

As long as the magnitude of the velocity and the radius of the turn do not change, the acceleration will remain constant.

Sally is to walk across a “high wire” that has been strung horizontally between two buildings that are apart. The sag (dip) in the rope when she stands at the midpoint is $8\degree$ . If her mass is , what is the tension in the rope at this point?

Explanation

The first thing is to identify the forces involved in this situation. There is the force of gravity (or her weight) which is pulling down on the rope. We can calculate this by

$F_G=(75kg)(-9.8\frac{m}{s^2})$

The other forces are the force of Tension on each side of the wire as she stands in the midpoint. These two Tension forces are what hold up Sally and keep her from falling. However, these two Tensions forces are at an $8\degree$ angle below the horizontal. This means that we need to analyze the components of the Tension force. The -components of each Tension force are equal in magnitude and opposite in direction as this is what keeps the rope connected to both buildings. The -components of each Tension force are equal in magnitude and in the same direction as they both are keeping Sally up. So we can sum up the forces acting in the -direction as:

$F_{net-Y} = F_G +F_{Tension-Y} + F_{Tension-Y}$

Which can be simplified to

$F_{net-Y} = F_G +2F_{Tension-Y}$

Since Sally is not accelerating, the forces are balanced and the net force must equal .

$0 = F_G +2F_{Tension-Y}$

Earlier we calculated the force of gravity so we can substitute this in to find the Tension in the Direction.

$0 = -735N +2F_{Tension-Y}$

$735N = 2F_{Tension-Y}$

$F_{Tension-Y} = 367.5N$

This is the Tension in the -direction. However, the problem is asking for the overall Tension in the wire. At this point, we must use trigonometric functions to determine the hypotenuse (the overall Tension) in the wire. Since the -component of the Tension is the opposite side of the triangle from the $8\degree$ angle, we can use cosine to find our hypotenuse.

$cos(\Theta ) = \frac{opp}{hyp}$

$cos(8) = \frac{367.5N}{hyp}$

$hyp = \frac{367.5N}{cos(8)}$

Two asteroids, one with a mass of $7.12*10^{18}kg$ and the other with mass , are $10*10^{10}m$ apart. What is the gravitational force on the LARGER asteroid?

$\small G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}$

$2.53*10^{-5}N$

$4.74*10^{-6}N$

$3.55*10^{-6}N$

$4.61*10^{-10}N$

Explanation

To solve this problem, use Newton's law of universal gravitation:

$F_G=G\frac{m_1m_2}{r^2}$

We are given the constant, as well as the asteroid masses and distance (radius). Using these values we can solve for the force.

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2}( \frac{(7.12*10^{18}kg)(5.33*10^{8}kg)}{(10*10^{10}m)^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (\frac{3.79*10^{27}kg}{10*10^{21}m^2})$

$F_G=6.67*10^{-11}\frac{m^3}{kg\cdot s^2} (3.79*10^5\frac{kg^2}{m^2})$

$F_G=2.53*10^{-5}N$

A crate slides across a floor for before coming to rest from its original position.

What is the coefficient of kinetic friction on the crate?

$\small g=-9.8\frac{m}{s^2}$

Explanation

The equation for the force due to friction is $F_f=\mu F_N$ , where $\mu$ is the coefficient of kinetic friction. Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation . We can equate these two force equations, meaning that $ma=\mu F_N$ . We can solve for the normal force, but we need to find in order to find $\mu$ .

The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

$\Delta x=\frac{(v_f+v_i)}{2}\Delta t$

Plug in our given values and solve.

$10m=\frac{(0\frac{m}{s}+v_i)}{2}5s$

$\frac{10m}{5s}=\frac{v_i}{2}$

$2\frac{m}{s}=\frac{v_i}{2}$

$4\frac{m}{s}=v_i$

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

$\Delta x=v_it+\frac{1}{2}at^2$

Plug in the given values to solve for acceleration.

$10m=(4\frac{m}{s}) (5s)+\frac{1}{2}a(5s)^2$

$10m=(20m)+\frac{1}{2}a(25s^2)$

$-10m=\frac{1}{2}a(25s^2)$

$\frac{-20m}{25s^2}=a$

$-0.8\frac{m}{s^2}=a$

Now that we have the acceleration and the mass, we can return to our first equation for force.

$F=ma=\mu F_N$

The normal force is the same as the mass times gravity.

$ma=\mu (mg)$

In this format, the masses cancel on both sides of the equation/

$a=\mu * (g)$

Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.

$-0.8\frac{m}{s^2}=\mu(-9.8\frac{m}{s^2})$

$\frac{-0.8\frac{m}{s^2}}{-9.8\frac{m}{s^2}}=\mu$

$0.082=\mu$

A box is released on a 25 degree incline and accelerates down the ramp at . What is the coefficient of kinetic friction impeding its motion?

Explanation

Consider the net forces acting on the object causing it to accelerate.

$F_{ne}t = F_{Gxdirection}-F_{friction}$

To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction. Using trigonometric functions we get that

$F_{GXdirection} = F_Gsin(\theta )$

We know that the force of gravity is equal to mg

$F_{GXdirection} = mgsin(\theta )$

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

$ma = mgsin(\theta )-F_{friction}$

The force of friction is directly related to μ (the coefficient of friction) times the normal force. In this case the normal force is equal to the y component of the force of gravity.

$F_{GYdirection} = mgcos(\theta )$