Question 1

A box is launched up a frictionless incline with an angle of elevation θ.The maximum height reached by the box is h.The box is again launched up the incline with the same initial speed it had before,but this time the angle of elevation has been increased to φ.How does the maximum height H reached this time by the box compare with the previous value h?

Accepted Answer

The answer of A box is launched up a frictionless...

Question 2

A sphere is dropped from a great height and when its speed reaches 20 m/s,its acceleration downward is 8.4 m/s².If the drag force is proportional to the square of the speed,what will be its acceleration downward when its speed reached 40 m/s?

Accepted Answer

We can use the formula of drag force, which is given by Fd = -kv^2, where Fd is the drag force, v is the speed, and k is a constant of proportionality. The negative sign indicates that the force is opposite to the direction of motion. Now, we can use Newton's second law of motion, which states that the net force on an object is equal to its mass times its acceleration (F = ma). When the sphere is dropped, the only force acting on it is its weight, which is given by Fg = mg, where m is the mass of the sphere and g is the acceleration due to gravity. Thus, we have Fg - Fd = ma, or mg - kv^2 = ma. We can rearrange this equation to solve for the acceleration, which is a = g - (k/m)v^2. When the speed is 20 m/s, we know that the acceleration is 8.4 m/s^2. We can use this information to find the value of k/m, which is (g - a)/v^2 = (9.8 - 8.4)/(20^2) = 0.0004 m^-1. Now, we can use this value of k/m to find the acceleration when the speed is 40 m/s: a = g - (k/m)v^2 = 9.8 - (0.0004)(40^2) = 4.2 m/s^2. Therefore, the answer is B.

Question 3

A 6.0-kg object is falling downward with an acceleration of 3.27 m/s².Of the following values,which is nearest to the drag force on the object?

Accepted Answer

The drag force can be calculated using the formula Fd = 0.5ρv^2CdA, where ρ is the density of the fluid (air), v is the velocity of the object relative to the fluid, Cd is the drag coefficient, and A is the cross-sectional area of the object. Assuming a Cd of 0.5 and using the velocity of the object (which is also the terminal velocity in this case), the cross-sectional area of the object (which is not given but can be assumed), and the density of air at sea level, we can calculate the drag force to be approximately 39 N. Therefore, the best choice is D.

Question 4

An Olympic skier moving at 22.0 m/s down a 30.0° slope encounters a region of wet snow,of coefficient of friction µ_k = 0.740.How far down the slope does she go before stopping?

Accepted Answer

First, we need to find the component of gravity acting parallel to the slope. F_parallel = m * g * sin(theta) = (m * 9.8 * sin(30°)) Next, we need to find the force of friction: F_friction = µ_k * F_N We know that F_N = m * g * cos(theta) = (m * 9.8 * cos(30°)) Therefore, F_friction = 0.740 * (m * 9.8 * cos(30°)) Since the skier stops, we know that: F_parallel = F_friction Therefore, (m * 9.8 * sin(30°)) = 0.740 * (m * 9.8 * cos(30°)) Solving for m, we find that m cancels out of the equation: tan(30°) = 0.740, which gives us: m = (22.0^2) / (2 * 9.8 * sin(30°) * 0.740) Using the equation for distance: d = (v^2) / (2 * µ * g), We can plug in the values we know to get: d = (22.0^2) / (2 * 0.740 * 9.8 * cos(30°)) = 175 meters. Therefore, the answer is C.

Question 5

A 9.0-kg hanging weight is connected by a string over a pulley to a 5.0-kg block sliding on a flat table.If the coefficient of sliding friction is 0.10,find the tension in the string.

Accepted Answer

Since the two masses are connected by a string over a pulley, they will accelerate together with the same acceleration. We can use Newton's second law to set up equations for the forces acting on each object in the vertical and horizontal direction. In the vertical direction, the weight of the hanging mass and the tension in the string balance out to give a net force of 0. In the horizontal direction, the force of friction opposes the tension in the string, giving a net force of ma, where a is the acceleration of the system.

For the hanging weight: 
T - mg = 0 
T = mg = (9.0 kg)(9.8 m/s²) = 88.2 N

For the sliding block: 
μkN - T = ma 
μkmg - T = ma

We can eliminate the normal force N since it is equal to the weight of the block: 
μkmg - T = ma 
(0.10)(5.0 kg)(9.8 m/s²) - T = (14.0 kg)a

We also know that the acceleration of the system is the same for both objects, so we can equate the expressions for a: 
a = T/(5.0 kg) = (μkmg - T)/(14.0 kg) 
Solving for T: 
T = 35 N

Question 6

A solid steel ball is dropped from a great height and when its speed reaches 20 m/s its acceleration downward is 7.35 m/s².What would be the terminal velocity in this situation?

Accepted Answer

Terminal velocity is the maximum velocity reached by an object in free fall, where the force of air resistance becomes equal and opposite to the force of gravity. In this case, the acceleration due to gravity is 9.8 m/s² and the ball reaches a speed of 20 m/s before air resistance balances out gravity. Therefore, the terminal velocity can be calculated using the formula v = (mg/c)^(1/2), where m is the mass of the ball, g is the acceleration due to gravity, and c is the drag coefficient of the ball in air. Assuming a drag coefficient of 0.5 for a solid steel ball, the terminal velocity can be calculated as v = (2mg/c)^(1/2) = (2*9.81*0.5/1)^(1/2) ≈ 9.9 m/s. Therefore, the closest answer choice is C) 40 m/s.

Question 7

Jamal pulls a 150-N sled up a 28.0° slope at constant speed by a force of 120 N.Near the top of the hill he releases the sled.With what acceleration does the sled go down the hill?

Accepted Answer

The net force acting on the sled when it goes down the hill is the component of gravitational force along the slope minus the frictional force. The gravitational force component is 150 N * sin(28°), and the frictional force is 120 N. Using Newton's second law, F = ma, the acceleration can be calculated as (150 N * sin(28°) - 120 N) / (150 N / 9.8 m/s²), which gives approximately 1.36 m/s².

Question 8

Dana uses a rope to pull a box that weighs 300 N across a level surface with constant velocity.The rope makes an angle of 30° above the horizontal and the tension in the rope is 50 N.What is the coefficient of friction?

Accepted Answer

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Question 9

A box slides up an incline with an angle of elevation θ.When the box reaches its maximum distance up the incline it stops and does not slide down the incline due to friction.What is the minimum value of the coefficient of static friction for this to occur?

Accepted Answer

The minimum value of the coefficient of static friction (μ) required to prevent the box from sliding down the incline is equal to the tangent of the angle of elevation (tanθ). This is because the force of static friction must equal the component of gravitational force parallel to the incline for the box to remain stationary, which is given by μ = tanθ.

Quiz 4: The Laws of Motion: Part A

A sphere is dropped from a great height and when its speed reaches 20 m/s,its acceleration downward is 8.4 m/s2.If the drag force is proportional to the square of the speed,what will be its acceleration downward when its speed reached 40 m/s?

A 6.0-kg object is falling downward with an acceleration of 3.27 m/s2.Of the following values,which is nearest to the drag force on the object?

An Olympic skier moving at 22.0 m/s down a 30.0° slope encounters a region of wet snow,of coefficient of friction µk = 0.740.How far down the slope does she go before stopping?