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MCAT Physical : Forces

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Example Question #11 : Friction And Normal Force

A 3kg book slides towards the right along a frictionless horizontal surface with initial velocity 5m/s, then suddenly encounters a long rough section with kinetic friction coefficient $\mu=0.25$ . How far does the book travel along the rough surface before coming to rest? (Use $\small \small g = 9.8\frac{m}{s^{2}}$ as needed)

Possible Answers:

4.4m

3.7m

6.5m

5.1m

2.9m

Correct answer:

5.1m

Explanation:

We'll need to use the kinematic equation $\small v_{f}^{2}=v_{i}^{2}+2ad$ to solve for d, the distance travelled when the book has stopped ( $\small v_{f}=0$ ). Before solving for d, we need to calculate the acceleration caused by the frictional force, by using the following steps.

1) Find the normal force on the book, $\small F_{n}=mg$ .

2) Plug this normal force into $\small F_{f}=\mu_{k}F_{n}$ to solve for frictional force.

3) Find the acceleration caused by this frictional force, with $\small F_{f}=ma$ .

Step 1 gives $\small \small F_{n}=29.4N$ , so in step 2, $\small \small F_{f}=7.35N$ , giving an acceleration of $\small a = 2.45\frac{m}{s^{2}}$ to the left (which we will define to be the negative horizontal direction).

Returning to the original kinematic equation, $\small 0 =(5 \frac{m}{s})^{2} + 2(-2.45\frac{m}{s^{2}})d$ .

Rearranging to solve for d gives $\small d = 5.1m$

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Example Question #11 : Friction And Normal Force

Which of the following could not influence the magnitude of frictional force acting on a book sliding across a horizontal table?

Possible Answers:

A pushing force exerted horizontally on the book

The mass of the book

The normal force of the table on the book

A pushing force exerted on the book at $\small 30^o$ below the horizontal

The materials of which the table and book are made

Correct answer:

A pushing force exerted horizontally on the book

Explanation:

Frictional force is given by the equation:

$F_{f}=\mu F_{N}$

The only factors that can change frictional force are the coefficient of friction, which is determined by the materials of the surfaces in contact, and the normal force. Since the normal force must have a magnitude such that the sum of forces perpendicular to the table equals zero, both the mass of the book and any external vertical forces would influence the normal force, and thus also would influence the frictional force.

$F_f=\mu (mg\sin(\theta))$

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Example Question #13 : Friction And Normal Force

A $\small 100g$ block rests on a wooden table. A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. The block just begins to move when the spring scale reads $\small 0.30N$ . What is the coefficient of static friction?

Possible Answers:

0.297

0.694

0.306

0.98N

0.306N

Correct answer:

0.306

Explanation:

The coefficient of friction is a ratio of the force initiating horizontal displacement of an object and the downwards force of the object due to gravity

$\small \mu_{s} = \frac{F_{pull}}{F_g}=\frac{0.30 N} {mg}$

$\mu_s=\frac{0.30N}{(0.1kg)(9.8\frac{m}{s^2})}=0.306$

Note that the units cancel since we are finding the ratio of two forces. Although it is possible to have values for the coefficient of static friction greater than one, these systems are highly unusual.

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Example Question #14 : Friction And Normal Force

A $\small 100g$ block rests on a wooden table with a coefficient of kinetic friction equal to $\small 0.27$ . A spring scale attached to the right side of the block is very gently pulled to the right with increasing force. If the spring scale is pulled with a force of $\small 0.62N$ , what is the acceleration of the system?

Possible Answers:

$1.78\frac{m}{s^2}$

$0.27\frac{m}{s^2}$

$4.85\frac{m}{s^2}$

$3.55\frac{m}{s^2}$

$0.62\frac{m}{s^2}$

Correct answer:

$3.55\frac{m}{s^2}$

Explanation:

The force of kinetic friction is given by the equation:

$F_f=\mu_kF_N=\mu_k(mg)$

We can calculate the frictional force using the values from the question.

$F_f=(0.27)(0.1kg)(-9.8\frac{m}{s^2})$

F_f=-0.265N

There are four force acting on the block: force from gravity, normal force, force of the spring scale, and force of friction. The normal force will be equal and opposite to the force of gravity, allowing us to cancel the forces in the vertical direction. This leaves us with a net force calculation for the horizontal force: the spring scale force and the force of friction. Note that the frictional force remains negative, as it acts in the opposite direction to the spring scale force.

$F_{net}=F_{ss}+F_f$

$F_{net}=0.62N+(-0.265N)=0.355N$

Now that we know the net force and the mass of the block, we can calculate the acceleration using Newton's second law.

F=ma

0.355N=(0.1kg)a

$a=\frac{0.355N}{0.1kg}=3.55\frac{m}{s^2}$

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Example Question #15 : Friction And Normal Force

What is the coefficient of kinetic friction of a 500g book sliding along a floor if the force of friction on the book is 4N?

Possible Answers:

1.25

0.8

0.25

0.125

Correct answer:

0.8

Explanation:

For formula for the force of friction is $F_k = \mu_{k}F_N$ . We can rearrange this equation to solve for the coefficient of friction.

$\mu_{k} = \frac{F_{k}}{F_{N}} = \frac{F_{k}}{mg}$

Remember that the normal force is equal to the force of gravity. Now we can plug in our given values and solve.

$\mu_k= \frac{4N}{(0.5kg)(10\frac{m}{s^2})}= 0.8$

Don't forget to convert 500g to 0.5kg. The units cancel out, leaving the answer without any unit.

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Example Question #1 : Understanding Friction

A 2kg box is at the top of a ramp at an angle of 60^o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released.

Imagine that the net force on the box is 16.5N when sliding down the ramp. What is the coefficient of kinetic friction for the box?

Possible Answers:

$\mu_k=0.16$

The kinetic coefficient of friction cannot be determined while the box is moving

$\mu_k=0.08$

$\mu_k=0.04$

Correct answer:

$\mu_k=0.08$

Explanation:

Since the box is moving when the net force on the box is determined, we can calculate the coefficient of kinetic friction for the box. The first step is determining what the net force on the box would be in the absence of friction. The net force on the box is given by the equation $\small F = mgsin\Theta$ .

$\small F = (2kg)(10\frac{m}{s^{2}})sin(60^{\circ}) = 17.3N.$

The difference between the frictionless net force and the net force with friction is 0.8N. This means that the force of kinetic friction on the box is 0.8N, acting opposite the direction of motion. Knowing this, we can solve for the coefficient of kinetic friction using the equation $\small f_{k} = \mu_{k}F_{n}$

$\small 0.8 = (mgcos\Theta)(\mu _{k})$

$\small 0.8 = (2kg)(10cos(60^{\circ}))\mu_{k}$

$\small \mu_{k} = 0.08$

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Example Question #1 : Universal Gravitation

Two objects with masses of M and m, sit r distance apart. What will be the effect on the gravitational force between them if the masses are changed to 2M and 3m?

It will increase 36-fold.
It will increase 6-fold.
It will increase 3-fold.
It will not change unless r is changed.
None of the above.

Possible Answers:

Correct answer:

Explanation:

Choice 2 is correct. The formula for gravitational force, G, is $\dpi{100} \small G=\frac{k\left ( m_{1} \right )\left ( m_{2} \right )}{r^{2}}$ , where k is a constant. The effect of doubling one mass and trebling the other is multiplicative, $\dpi{100} \small 2\times 3$ , so the answer is six-fold. The question attempts to confuse the respondent by forcing them to recall that the element of the equation which is squared is distance between objects.

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Example Question #2 : Universal Gravitation

An object has a mass of 50kg and a weight of 500N when it is resting on the surface of the Earth. If it is moved to a height equal to three times the Earth’s radius, what is the object’s new weight?

Possible Answers:

30N

60N

130N

300N

Correct answer:

30N

Explanation:

You should be familiar with the following equation for the force of gravity.

$F_g = G\frac{m_{1}m_{2}}{r^{2}}$

To solve this problem, recognize that weight (F_g) is proportional to the inverse square of the radius. When the object was a distance of r (Earth’s radius) it had a weight of 500N. Now, the object is at a distance of 4r (radius of Earth plus the 3r distance that the object is moved to). With the proportion described above, we can see that the force is decreased by a factor of (4)².

$\frac{1}{4^{2}}(500 N) = \frac{1}{16}(500 N) = 30 N$

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Example Question #1 : Universal Gravitation

The moon's distance from the center of the Earth was decreased by a multiple of three. How would this affect the gravitational force of the Earth on the moon?

Possible Answers:

The gravitational force would not be affected, because the moon's mass is the same

It would increase by a factor of three

It would decrease by a factor of nine

It would increase by a factor of nine

Correct answer:

It would increase by a factor of nine

Explanation:

The law of gravitation is written as $\small F = \frac{Gm_{1}m_{2}}{r^{2}}$ , with G being equal to $\small \small 6.67*10^{-11}\frac{m^3}{kg*s^2}$ .

Since the radius of the two masses acting on each other is squared, and is found in the denominator, a decrease in the radius by a multiple of three will cause a nine-fold increase in the gravitational force.

$\small F_1 = \frac{Gm_{1}m_{2}}{r^{2}}\rightarrow F_2=\frac{Gm_{1}m_{2}}{(\frac{1}{3}r)^{2}}=9F_1$

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Example Question #4 : Universal Gravitation

A body with mass $\small m$ is situated $\small 3.0$ meters from a second body with a mass of $\small 2m$ . What will be the effect on gravitational attraction of moving one body so that it is only $\small 1.74$ meters from the other body?

Possible Answers:

It will increase by a factor of $\small 9$

It will increase by a factor of $\small 3$

It will decrease by a factor of $\small 1.74$

It will decrease by a factor of $\small 3$

It will increase by a factor of $\small 1.74$

Correct answer:

It will increase by a factor of $\small 3$

Explanation:

Gravity is essentially a property of mass, and the force of gravitational attraction between two bodies is given by the formula:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

In our scenario, the masses remain the same, and of course $\small G$ is a constant, so the only thing that changes is the denominator.

$F_{g1} = G\frac{m_{1}m_{2}}{(3.0)^{2}}=G\frac{m_{1}m_{2}}{9}$

$F_{g2} = G\frac{m_{1}m_{2}}{(1.74)^{2}}=G\frac{m_{1}m_{2}}{3.03}$

$\frac{9}{3.03} \cong 3$

$F_{g2}\approx3F_{g1}$

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MCAT Physical : Forces

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #11 : Friction And Normal Force

Example Question #11 : Friction And Normal Force

Example Question #13 : Friction And Normal Force

Example Question #14 : Friction And Normal Force

Example Question #15 : Friction And Normal Force

Example Question #1 : Understanding Friction

Example Question #1 : Universal Gravitation

Example Question #2 : Universal Gravitation

Example Question #1 : Universal Gravitation

Example Question #4 : Universal Gravitation

All MCAT Physical Resources

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