Forces

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MCAT Physical › Forces

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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)

5.1m

4.4m

3.7m

2.9m

6.5m

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.

Find the normal force on the book, $\small F_{n}=mg$ .
Plug this normal force into $\small F_{f}=\mu_{k}F_{n}$ to solve for frictional force.
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 $\small d = 5.1m$

A 50kg block resting on the ground experiences an upward acceleration of 4m/s2. What is the normal force acting on the block?

300 N

200 N

700 N

500 N

Explanation

The block experiences the force of gravity, plus the force of the upward acceleration

$Fnet = (50kg)(-10m/s^{2}) + (50kg)(4m/s^{2}) = -300 N$

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.

Which of the following scenarios describes a system that is not in equilibrium?

A racecar driver driving at $200\frac{m}{s}$ around a mile-long circular track

A tightrope walker balancing on one foot on a wire

A skydiver that has deployed his parachute and is falling straight down at $25\frac{m}{s}.$

All of these scenarios describe systems in equilibrium

Explanation

The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).

Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.

When the force applied to a moving object is equal and opposite the force of kinetic friction, what happens to the object?

It moves with a constant velocity

It stops

It accelerates

It decelerates

It changes direction

Explanation

It is important to understand the difference between static and kinetic friction. When an object is at rest, it takes more force to get it to start moving than to keep it moving. If you match the amount of static friction that can be generated when the object is at rest, it will not move because there is zero net force; the force applied must be greater than the static friction in order to initiate motion. Once the object begins moving, the force required to keep it moving decreases. If you match the force of kinetic friction, the object moves at a constant velocity because there is again no net force. Any more force will cause acceleration, while any less will cause deceleration.

A force of is applied to a block with a mass of on a frictionless surface. What is the acceleration of the block?

$4\frac{m}{s^{2}}$

$5\frac{m}{s^{2}}$

$100\frac{m}{s^{2}}$

The block moves with a constant velocity

Explanation

For this question we must use Newton's second law. We have to use the formula .

is the force being applied, is the mass of the object, and is the acceleration of the object

$20N = 5kg \cdot a$

$a = \frac{20N}{5kg}$

$a = 4 \frac{m}{s^{2}}$

A 2kg box is at the top of a ramp at an angle of 60o. 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?

$\mu_k=0.08$

$\mu_k=0.16$

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

$\mu_k=0.04$

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 $\small f_{k} = \mu_{k}F_{n}$

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

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

$\small \mu_{k} = 0.08$

When the force applied to a moving object is equal and opposite the force of kinetic friction, what happens to the object?

It moves with a constant velocity

It stops

It accelerates

It decelerates

It changes direction

Explanation

5.1m

4.4m

3.7m

2.9m

6.5m

Explanation

Find the normal force on the book, $\small F_{n}=mg$ .
Plug this normal force into $\small F_{f}=\mu_{k}F_{n}$ to solve for frictional force.
Find the acceleration caused by this frictional force, with $\small F_{f}=ma$ .

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 $\small d = 5.1m$

A 50kg block resting on the ground experiences an upward acceleration of 4m/s2. What is the normal force acting on the block?

300 N

200 N

700 N

500 N

Explanation

The block experiences the force of gravity, plus the force of the upward acceleration

$Fnet = (50kg)(-10m/s^{2}) + (50kg)(4m/s^{2}) = -300 N$

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.

A force of is applied to a block with a mass of on a frictionless surface. What is the acceleration of the block?

$4\frac{m}{s^{2}}$

$5\frac{m}{s^{2}}$

$100\frac{m}{s^{2}}$

The block moves with a constant velocity

Explanation

For this question we must use Newton's second law. We have to use the formula .

is the force being applied, is the mass of the object, and is the acceleration of the object

$20N = 5kg \cdot a$

$a = \frac{20N}{5kg}$

$a = 4 \frac{m}{s^{2}}$

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