MCAT Physical : Newtonian Mechanics and Motion
Study concepts, example questions & explanations for MCAT Physical
All MCAT Physical Resources
Example Questions
Example Question #3 : Forces
Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
When the force of friction acts on a moving body, as it would on Sam when moving down the hill, __________.
Entropy of the universe decreases because movement decreases as Sam slows to a stop
Entropy increases in the universe
Friction always acts opposite the direction of movement
Friction is inversely proportional to the normal force
Friction can be totally recaptured and used to power future movements
Entropy increases in the universe
This is a tricky question. Entropy of the universe increases because, even though Sam's movement does slow down resulting in a local decline in entropy, the heat generated by friction with the ground results in a net increase in entropy overall.
Another tempting choice would be that friction always acts opposite the direction of motion. Friction always acts opposite to the direction of RELATIVE motion, not necessarily to the direction of motion in which the body itself is traveling.
Example Question #4 : Forces
Two children are playing with sleds on a snow-covered hill. Sam weighs 50kg, and his sled weighs 10kg. Sally weighs 40kg, and her sled weighs 12kg. When they arrive, they climb up the hill using boots. Halfway up the 50-meter hill, Sally slips and rolls back down to the bottom. Sam continues climbing, and eventually Sally joins him at the top.
They then decide to sled down the hill, but disagree about who will go first.
Scenario 1:
Sam goes down the hill first, claiming that he will reach a higher velocity. If Sally had gone first, Sam says they could collide.
Scenario 2:
Sally goes down the hill first, claiming that she will experience lower friction and thus reach a higher velocity. If Sam had gone first, Sally says they could collide.
Scenario 3:
Unable to agree, Sam and Sally tether themselves with a rope and go down together.
What would you expect is true of Sam before he starts moving?
There would be more friction acting on Sam while he moved down the hill if he weighed less, since he would be traveling more slowly
The maximum force of friction acting on Sam is greater before Sam starts moving than once he is sliding down the hill
Friction can only act on a body while it is slowing down
The maximum force of friction acting on Sam is less before Sam starts moving than once he is sliding down the hill
There can only be friction acting on Sam when he is in motion
The maximum force of friction acting on Sam is greater before Sam starts moving than once he is sliding down the hill
Static friction exerts a greater effect than does kinetic friction. Think about if you are trying to slide a heavy box across a room. It takes some force to get it going, but once it is moving it takes comparatively less force to keep it moving
Example Question #1 : Forces
Two children are playing on an icy lake. Child 1 weighs 50kg, and child 2 weighs 38kg. Child 1 has a backpack that weighs 10kg, and child 2 has a backpack that weighs 5kg.
Over the course of the afternoon, they collide many times. Four collisions are described below.
Collision 1:
Child 1 starts from the top of a ramp, and after going down, reaches the lake surface while going 5m/s and subsequently slides into a stationary child 2. They remain linked together after the collision.
Collision 2:
Child 1 and child 2 are sliding in the same direction. Child 2, moving at 10m/s, slides into child 1, moving at 2m/s.
Collision 3:
The two children collide while traveling in opposite directions at 10m/s each.
Collision 4:
The two children push off from one another’s back, and begin moving in exactly opposite directions. Child 2 moves with a velocity of +8m/s.
Imagine that the above collisions happened on another planet, where the gravitational constant was exactly half that of Earth. Compared to the force of friction acting on moving bodies on Earth, the force of friction on this new planet would be __________.
less, because the normal force would decrease
more, because the normal force would decrease
the same, because the coefficients of friction are functions of material, and are independent of gravity
more, because the normal force would increase
less, because the normal force would increase
less, because the normal force would decrease
While the answer choice specifying the material-dependency of the coefficient of friction is true, and may be tempting, the force of friction is dependent on both the coefficients of friction and the normal force. The normal force is directly proportional to the weight of the body in question.
Example Question #5 : Friction And Normal Force
A crate of mass M is pushed along a frictionless horizontal floor by a force F acting downward from an angle xo above the horizontal. What is the magnitude of the normal force N between the crate and the floor?
Since the crate is not accelerating in the vertical direction, the sum of the vertical forces must be 0. In other words, the sum of the downwards forces must equal the sum of the upwards forces. In this case, the downwards forces on the crate are gravity (Mg) and the vertical component of the pushing force (Fsin(x)). The only force acting upwards is the normal force (N) from the floor. So, Mg + Fsin(x) = N. Note that the problem states the direction of the force to be from above the horizontal, meaning that the vertical component will add to the downward force.
Example Question #1 : Forces
A 50kg block resting on the ground experiences an upward acceleration of 4m/s2. What is the normal force acting on the block?
200 N
700 N
500 N
300 N
300 N
The block experiences the force of gravity, plus the force of the upward acceleration
If the block is resting on the ground, then its total force must be zero, and the normal force must cancel out the net force above. The normal force is 300N.
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 
3.7m
2.9m
4.4m
6.5m
5.1m
5.1m
We'll need to use the kinematic equation 
1) Find the normal force on the book, 
2) Plug this normal force into 
3) Find the acceleration caused by this frictional force, with 
Step 1 gives 
Returning to the original kinematic equation, 
Rearranging to solve for d gives
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?
The mass of the book
A pushing force exerted on the book at 
The normal force of the table on the book
The materials of which the table and book are made
A pushing force exerted horizontally on the book
A pushing force exerted horizontally on the book
Frictional force is given by the equation:
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.
Example Question #12 : Friction And Normal Force
A 
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
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.
Example Question #1 : Gre Subject Test: Physics
A 
The force of kinetic friction is given by the equation:
We can calculate the frictional force using the values from the question.
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.
Now that we know the net force and the mass of the block, we can calculate the acceleration using Newton's second law.
Example Question #11 : Forces
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?
For formula for the force of friction is 
Remember that the normal force is equal to the force of gravity. Now we can plug in our given values and solve.
Don't forget to convert 500g to 0.5kg. The units cancel out, leaving the answer without any unit.
Certified Tutor
Certified Tutor
All MCAT Physical Resources
