Dynamics Review

If an object of mass m is on a ramp of angle θ, the magnitude of the component of gravity parallel to the ramp is ​ ​ (a)   and the magnitude of the component of gravity perpendicular to the ramp is ​ (b)   .

Answer explanation

The component of gravity parallel to the ramp is given by mg sin(θ), while the component perpendicular to the ramp is mg cos(θ). Thus, the correct answers are mg sin(θ) and mg cos(θ).

Tags

NGSS.HS-PS2-1

When Mia and Samuel are pushing a box up an inclined plane, what happens to the normal force as the angle of the incline increases?

• The normal force increases

• The normal force decreases

• The normal force is not affected by the incline

• The normal force is only dependent on the force due to gravity

Answer explanation

As the angle of the incline increases, the component of gravitational force acting perpendicular to the incline decreases, leading to a reduction in the normal force. Therefore, the normal force decreases.

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4.3 (2nd Law)

A box of mass m slides up a ramp with initial velocity +v₀. The kinetic friction force on the box has magnitude F. Which of the following is a correct equation that could be used to determine the acceleration a of the box?

Answer explanation

The correct equation is -F - mg sin(θ) = ma. This accounts for the kinetic friction force (F) and the component of gravitational force acting down the ramp (mg sin(θ)), both opposing the box's motion.

Tags

NGSS.HS-PS2-1

An object sits on level ground and experiences an applied force with magnitude F_A. Place the following scenarios in order from the least magnitude of the normal force to the greatest magnitude of the normal force.

• 
• 
• 

Answer explanation

The key is noticing the direction on the applied force. Angled upward, the vertical component of the applied force plus the normal force cancels out gravity, making this the case with the least normal force. Angled directly to the right makes the normal force equal to the force of gravity. Angled downward, the normal force has to cancel out the vertical component of the applied force plus gravity. Therefore, this has the greatest normal force.

Tags

NGSS.HS-PS2-1

A box weighing 95 N experiences a net force of 41 N while it is being lifted. What is the acceleration of the box?

• 4.2 m/s²

• 0.27 m/s²

• 0.43 m/s²

• 2.2 m/s²

Answer explanation

To find the acceleration, use Newton's second law: F = ma. The net force is 41 N, and the mass can be calculated from weight (W = mg). Here, m = W/g = 95 N/9.8 m/s² ≈ 9.7 kg. Thus, a = F/m = 41 N / 9.7 kg ≈ 4.2 m/s².

Tags

NGSS.HS-PS2-1

An 6.0 kg object on a foreign planet is in freefall. After dropping from rest, it accelerates to 12 m/s in 3.0 seconds. Assume wind resistance is negligible. What is the weight of the object in Newtons?

• 6 N

• 24 N

• 60 N

• 18 N

Answer explanation

The weight of an object is calculated using the formula: weight = mass × gravity. Here, mass = 6.0 kg and gravity can be derived from acceleration (12 m/s in 3s gives 4 m/s²). Thus, weight = 6.0 kg × 4 m/s² = 24 N.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4

A 2 kg block is pushed across a surface with a force of 10 N. If the coefficient of kinetic friction is 0.2, what is the net force acting on the block? (Use g = 10 m/s²)

• 2 N

• 4 N

• 6 N

• 8 N

Answer explanation

First, calculate the frictional force: F_friction = μ * m * g = 0.2 * 2 kg * 10 m/s² = 4 N. The net force is then F_net = applied force - friction = 10 N - 4 N = 6 N. Thus, the correct answer is 6 N.

Tags

NGSS.HS-PS2-1

An object is on a ramp, and there is static friction between the object and the ramp keeping it in place. Which of the following is INCORRECT about the free body diagram to the left?

• The friction vector is pointed in the opposite direction.

• The gravity vector should be pointed diagonally in the direction of the ramp.

• The normal force vector should be at an angle perpendicular to the ramp.

• There is nothing incorrect about the free body diagram.

• There should be an applied force.

Answer explanation

The normal force vector should always be directed perpendicular to the surface. Since this surface is angled, the normal force vector should also be angled.

Tags

NGSS.HS-PS2-1

Mathematical Equivalence

ON

Answer explanation

Setting up F = ma and noticing that the gravity on the larger block is in the direction of motion and the gravity on the smaller block is against the direction of motion, we get (15)(10) - (5)(10) = (5+15)a. This means a = (150 - 50)/20 = 5 meters per second squared.

Tags

NGSS.HS-PS2-1

Mathematical Equivalence

ON

Answer explanation

Taking a = 5 m/s^2, we can isolate one of the blocks to find the tension. Isolating the larger block and taking the direction of motion (down) to be positive, we get (15)(10) - T = (15)(5), so T = 150 - 75 = 75 N.

Tags

NGSS.HS-PS2-1

NGSS.HS-PS2-4