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High School Physics : High School Physics

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6 Diagnostic Tests 233 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Specific Forces

Susan is trying to push a 12kg crate across the floor. She observes that the force of friction between the crate and the floor is -50N . What is the coefficient of static friction?

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

Possible Answers:

0.11

0.43

0.56

0.23

It cannot be determined

Correct answer:

0.43

Explanation:

The equation for the force of friction is $F_{friction}=\mu F_{normal}$ , where $\mu$ is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).

$F_{normal}=-F_{gravity}=-(mg)$

$F_{normal}=-((12kg)(-9.8\frac{m}{s^2}))$

$F_{normal}=-(-117.6N)$

$F_{normal}=117.6N$

Since the problem tells us that the force due to friction is -50N , we can plug these values into our original equation to solve for the coefficient of friction.

$F_{friction}=\mu F_{normal}$

$(-50N)=\mu (117.6N)$

$\frac{-50N}{117.6N}=\mu$

$0.43=\mu$

The coefficient of friction has no units.

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Example Question #2 : Specific Forces

Mary is trying to push a 12kg crate across the floor. She observes that the coefficient of static friction between the floor and the crate is equal to 0.12 . What will be the force of friction on the crate?

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

Possible Answers:

-11.23N

-0.32N

-9.81N

-14.11N

Correct answer:

-14.11N

Explanation:

The equation for the force of friction is $F_{friction}=\mu F_{normal}$ , where $\mu$ is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative the force of gravity).

$F_{normal}=-F_{gravity}=-(mg)$

$F_{normal}=-((12kg)(-9.8\frac{m}{s^2}))$

$F_{normal}=-(-117.6N)$

$F_{normal}=117.6N$

Since the problem tells us that the coefficient due to friction is equal to 0.12 , we can plug these values into our original equation to solve for the force.

$F_{friction}=\mu F_{normal}$

$F_{friction}=(0.12)(117.6N)$

$F_{friction}=14.11N$

As the force due to friction is moving in the OPPOSITE direction to the force Mary will exert, it should appropriately be labelled as negative.

$F_{friction}=-14.11N$

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Example Question #1 : Specific Forces

A 2.9kg crate slides across a floor for before coming to rest 10m from its original position.

What is the coefficient of kinetic friction on the crate?

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

Possible Answers:

0.004

0.098

1.6

0.082

0.8

Correct answer:

0.082

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 F=ma . 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.

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

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

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

-20m=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.

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

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

$0.082=\mu$

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Example Question #1 : Understanding Frictional Force

Erin pushes a 38kg cabinet across the floor. If it requires 171N of force, what is the coefficient of kinetic friction?

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

Possible Answers:

2.18

0.372

0.214

4.68

0.46

Correct answer:

0.46

Explanation:

The problem gives us the minimum force required to move the cabinet. That means the force Erin exerts will be equal to the force due to friction, but moving in the opposite direction.

F=-F_f .

From here, expand the right side using the formula for kinetic friction and normal force.

$F=-(\mu*F_N)$

$F=-\mu(mg)$

Use the given values for the force, mass, and acceleration of gravity to solve for the coefficient of friction.

$171N=-\mu(38kg*-9.8\frac{m}{s^2})$

$171N=-\mu(-372.4N)$

$\frac{171N}{-372.4N}=-\mu$

$-0.46=-\mu$

$0.46=\mu$

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Example Question #11 : Specific Forces

It takes 20N of force to move a 19kg block from rest. Assuming no outside forces act upon the block, what is the coefficient of static friction?

Possible Answers:

0.21

9.31

0.10

0.98

0.11

Correct answer:

0.11

Explanation:

The force due to friction on a level surface is the product of the normal force and the coefficient of friction:

$F_f=\mu_{s}F_N$

We are told the force required to move the block, which will be equal to the force of friction, and the mass of the block. Use the mass of the block to calculate the normal force.

F_N=mg

$F_N=(19kg)(9.8\frac{m}{s^2})=186.2N$

Use the normal force and the force of friction to solve for the coefficient of friction.

$F_f=\mu_{s}F_N$

$20N=\mu_s(186.2N)$

$\mu_s=\frac{20N}{186.2N}$

$\mu_s=0.11$

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Example Question #1 : Understanding Frictional Force

Which of the following cannot be true of an object on a given surface?

Possible Answers:

Static friction: 10N

Kinetic friction:

Static friction:

Kinetic friction:

Static friction:

Kinetic friction:

Static friction: 10N

Kinetic friction:

Static friction:

Kinetic friction:

Correct answer:

Static friction:

Kinetic friction:

Explanation:

Kinetic friction is never greater than static friction. More force is always requires to overcome static friction than is required to overcome kinetic friction. It can require a large force to initiate motion, causing an initial acceleration by overcoming static friction. Once motion has begun, however, less for is required to maintain the motion due to the principles of Newton's first law and inertia.

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Example Question #1 : Understanding Gravity And Weight

Michael lands on a new planet. If his mass is 20kg and the acceleration due to gravity on this planet is $-1.56\frac{m}{s^2}$ , what would his weight be?

Possible Answers:

-112.1N

-31.2N

-68.4N

-32.5N

-189.2N

Correct answer:

-31.2N

Explanation:

Weight is a particular force that is equal to mass times acceleration due to gravity. Start with Newton's second law, $\vec{F}=m\vec{a}$ .

$\vec{F}_{weight}=m\vec{a}_{gravity}$

Plug in the given values to solve.

$\vec{F}=(20kg)(-1.56\frac{m}{s^2})$

$\vec{F}_{weight}=-31.2N$

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Example Question #1 : Understanding Gravity And Weight

Leslie is an astronaut on a new planet. She knows her mass is 70kg , and she can calculate her weight on the planet to be -512N . What is the acceleration due to gravity on the new planet?

Possible Answers:

$-3.64\frac{m}{s^2}$

$-9.81\frac{m}{s^2}$

$-7.65\frac{m}{s^2}$

$-12.31\frac{m}{s^2}$

$-7.31\frac{m}{s^2}$

Correct answer:

$-7.31\frac{m}{s^2}$

Explanation:

Weight is a specific force. It is always equal to the mass times acceleration due to gravity: $\Vec{F}_{weight}=m\vec{a}_{gravity}$ .

Plug in the given values to solve for the acceleration.

$(-512N)=(70kg)(\vec{a})$

$\frac{-512N}{70kg}=\vec{a}$

$-7.31\frac{m}{s^2}=\vec{a}$

The gravitational acceleration is negative because it acts in the downward direction.

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Example Question #3 : Understanding Gravity And Weight

Jenny calculates that her weight on Earth is equal to -500N . Assuming acceleration due to gravity on earth is equal to $-9.8\frac{m}{s^2}$ , what is her mass?

Possible Answers:

9.82kg

51.02kg

101.2kg

5.10kg

53.23kg

Correct answer:

51.02kg

Explanation:

Weight is a specific force that is always equal to mass times acceleration due to gravity.

Start with Newton's second law: $\vec{F}=m\vec{a}$ .

$\vec{F}_{weight}=m\vec{a}_{gravity}$

Plug in the given values to solve for the mass.

$-500N=(m)(-9.8\frac{m}{s^2})$

$\frac{-500N}{-9.8\frac{m}{s^2}}=m$

51.02kg=m

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Example Question #4 : Understanding Gravity And Weight

Jerry wants to lift a 0.12kg ball with exactly enough force so that it's upward velocity is constant. How much force should he use?

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

Possible Answers:

2.36N

3.32N

1.18N

2.18N

Correct answer:

1.18N

Explanation:

If the velocity on an object is constant, that means it has no acceleration. If it has no acceleration, that means that the net force on the object is equal to zero. We can see this conclusion by using Newton's second law.

$F_{net}=ma$

$F_{net}=m(0\frac{m}{s^2})$

$F_{net}=0N$

Another way to think of $F_{net}$ is the sum of all the forces. Since the only two forces acting upon the ball are gravity and Jerry's lifting force, we can see: $F_{net}=F_{Jerry}+F_{gravity}$ .

Since the net force is zero, the magnitude of Jerry's force must equal the magnitude of the force of gravity, but in the opposite direction.

$0=F_{Jerry}+F_{gravity}$

$-F_{Jerry}=F_{gravity}$

This means that once we find $F_{gravity}$ , then Jerry's lifting force will be the same magnitude but in the opposite direction. Use Newton's second law to find the force of gravity.

$F_{gravity}=ma$

$F_{gravity}=(0.12kg)(-9.8\frac{m}{s^2})$

$F_{gravity}=-1.18N$

This means that Jerry's lifting force will be 1.18N .

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High School Physics : High School Physics

Study concepts, example questions & explanations for High School Physics

All High School Physics Resources

Example Questions

Example Question #1 : Specific Forces

Example Question #2 : Specific Forces

Example Question #1 : Specific Forces

Example Question #1 : Understanding Frictional Force

Example Question #11 : Specific Forces

Example Question #1 : Understanding Frictional Force

Example Question #1 : Understanding Gravity And Weight

Example Question #1 : Understanding Gravity And Weight

Example Question #3 : Understanding Gravity And Weight

Example Question #4 : Understanding Gravity And Weight

All High School Physics Resources

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