Friction and Inclined Planes Quiz

Which of the following is true about static friction?

• Static friction always opposes motion.

• Static friction is always greater than kinetic friction.

• Static friction exists only when an object is in motion.

• Static friction is independent of the normal force.

Answer explanation

Static friction is indeed always greater than kinetic friction, as it must overcome the initial resistance to motion. The other statements are incorrect: static friction opposes the start of motion, exists when at rest, and depends on the normal force.

The coefficient of kinetic friction is always greater than the coefficient of static friction.

• True

• False

Answer explanation

The statement is false because the coefficient of static friction is typically greater than the coefficient of kinetic friction. Static friction prevents motion, while kinetic friction acts when objects are already sliding.

An object of mass 10 kg is placed on an inclined plane with an angle of 30° to the horizontal. What is the component of the object's weight parallel to the surface of the incline? (Use g=9.8 m/s²)

• 98 N

• 49 N

• 85 N

• 19 N

Answer explanation

To find the weight component parallel to the incline, use W_parallel = mg sin(θ). Here, m = 10 kg, g = 9.8 m/s², and θ = 30°. Thus, W_parallel = 10 * 9.8 * sin(30°) = 49 N. Therefore, the correct answer is 49 N.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4

If the angle of the incline is increased, what happens to the frictional force acting on an object on the incline (assuming no change in the coefficient of friction)?

• It increases.

• It decreases.

• It remains the same.

• It becomes zero.

Answer explanation

As the angle of the incline increases, the component of gravitational force parallel to the incline increases, reducing the normal force. Since frictional force is proportional to the normal force, it decreases.

An object is dropped from rest from a height of 50 m. What is the velocity of the object just before it hits the ground?

Answer explanation

Using the formula v = √(2gh), where g = 9.81 m/s² and h = 50 m, we find v = √(2 * 9.81 * 50) ≈ 31.3 m/s. Thus, the velocity just before hitting the ground is approximately 31.3 m/s.

Tags

NGSS.HS-PS2-1

A car accelerates from rest at 2 m/s² for 10 seconds. How far does the car travel during this time?

Answer explanation

The distance traveled can be calculated using the formula: distance = 0.5 * acceleration * time². Here, distance = 0.5 * 2 m/s² * (10 s)² = 100 m. Thus, the car travels 100 meters.

Tags

NGSS.HS-PS2-1

A sled of mass 50 kg is pulled up a snow-covered hill that is inclined at 25° to the horizontal. The coefficient of kinetic friction between the sled and the snow is 0.1. Calculate the force required to pull the sled up the incline at a constant velocity. Show all steps in your solution.

Answer explanation

To find the force, calculate the gravitational force component down the incline (mg sin(25°)) and the frictional force (μN, where N = mg cos(25°)). The total force equals these two forces to maintain constant velocity.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4

Static friction is generally greater than kinetic friction.

• True

• False

Answer explanation

Static friction is the force that prevents motion between two surfaces in contact, while kinetic friction acts when they are sliding past each other. Static friction is typically greater than kinetic friction, making the statement true.

As the angle increases, the normal force decreases, and the frictional force decreases as a result.

• True

• False

Answer explanation

As the angle of an inclined plane increases, the component of gravitational force acting perpendicular to the surface (normal force) decreases. This reduction in normal force leads to a decrease in frictional force, making the statement true.

A ball is thrown upwards with an initial velocity of 15 m/s. How long will it take for the ball to reach the highest point of its trajectory?

Answer explanation

1. Time to reach the highest point: t=v0g=159.8≈1.53 secondst = \frac{v_0}{g} = \frac{15}{9.8} \approx 1.53 \, \text{seconds}t=gv0​​=9.815​≈1.53seconds.