Question 1

A person is dragging a packing crate of mass 100 kg across a rough horizontal floor where the coefficient of kinetic friction is 0.400.He exerts a force F sufficient to accelerate the crate forward.At what angle above horizontal should his pulling force be directed in order to achieve the maximum acceleration?

Accepted Answer

The correct choice is A because the coefficient of kinetic friction is 0.400, so the angle above horizontal should be 21.8° in order to achieve the maximum acceleration.

Question 2

Two blocks are connected by a string that goes over an ideal pulley as shown in the figure.Block A has a mass of 3.00 kg and can slide over a rough plane inclined 30.0° to the horizontal.The coefficient of kinetic friction between block A and the plane is 0.400.Block B has a mass of 2.77 kg.What is the acceleration of the blocks?

Accepted Answer

The answer of Two blocks are connected by a string...

Question 3

The figure shows two packages that start sliding down a 20° ramp from rest a distance  along the ramp from the bottom.Package A has a mass of 5.0 kg and a coefficient of kinetic friction 0.20 between it and the ramp.Package B has a mass of 10 kg and a coefficient of kinetic friction 0.15 between it and the ramp.How long does it take package A to reach the bottom?

Accepted Answer

First, we need to calculate the gravitational force acting on each package down the ramp. Using the formula F=mg(sinθ), where θ is the angle of the ramp (20°) and g is the acceleration due to gravity (9.8 m/s^2), we get:
- For package A: F=5.0 kg x 9.8 m/s^2 x sin(20°) = 16.9 N
- For package B: F=10 kg x 9.8 m/s^2 x sin(20°) = 33.8 N

Next, we need to calculate the force of kinetic friction acting on each package as it slides down the ramp. Using the formula f=μkN, where μk is the coefficient of kinetic friction and N is the normal force (equal to the component of the gravitational force perpendicular to the ramp), we get:
- For package A: f=0.20 x 5.0 kg x 9.8 m/s^2 x cos(20°) = 8.8 N
- For package B: f=0.15 x 10 kg x 9.8 m/s^2 x cos(20°) = 8.3 N

The net force acting on each package down the ramp is therefore:
- For package A: Fnet = F - f = 16.9 N - 8.8 N = 8.1 N
- For package B: Fnet = F - f = 33.8 N - 8.3 N = 25.5 N

We can now use the formula F=ma, where a is the acceleration of each package down the ramp, to solve for a:
- For package A: 8.1 N = 5.0 kg x a, so a = 1.62 m/s^2
- For package B: 25.5 N = 10 kg x a, so a = 2.55 m/s^2

Finally, we can use the formula x = 1/2at^2 + vt, where x is the distance traveled down the ramp (11 m), v is the initial velocity (0 m/s), and t is the time taken to travel the distance x, to solve for t:
- For package A: 11 m = 1/2 x 1.62 m/s^2 x t^2, so t = 2.68 s ≈ 2.7 s
- For package B: 11 m = 1/2 x 2.55 m/s^2 x t^2, so t = 2.79 s (this is not one of the answer choices, so we can eliminate B, C, and D)

Question 4

A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal.If the coefficient of kinetic friction is 0.19,at what rate does the box accelerate down the slope?

Accepted Answer

The force acting on the box down the slope can be calculated as:
F_downslope = m * g * sin(θ) = 6.0 kg * 9.81 m/s^2 * sin(39°) ≈ 37.3 N

The force of kinetic friction can be calculated as:
F_friction = μ * F_norm = μ * m * g * cos(θ) = 0.19 * 6.0 kg * 9.81 m/s^2 * cos(39°) ≈ 22.9 N

The net force acting on the box down the slope is:
F_net = F_downslope - F_friction = 37.3 N - 22.9 N ≈ 14.4 N

The acceleration of the box down the slope can be calculated as:
a = F_net / m = 14.4 N / 6.0 kg ≈ 2.4 m/s^2

Therefore, the rate at which the box accelerates down the slope is:
v = a * t ≈ 2.4 m/s^2 * t

However, the initial velocity of the box is not given, so we cannot determine the rate of acceleration without additional information. Therefore, none of the answer choices are correct.

Question 5

Three objects are connected as shown in the figure.The strings and frictionless pulleys have negligible masses,and the coefficient of kinetic friction between the 2.0-kg block and the table is 0.25.What is the acceleration of the 2.0-kg block?

Accepted Answer

The answer of Three objects are connected as shown in...

Question 6

In the figure,two wooden blocks each of 0.30 kg mass are connected by a string that passes over a very light frictionless pulley.One block slides on a horizontal table,while the other hangs suspended by the string,as shown in the figure.At time t = 0,the suspended block is 0.80 m over the floor,and the blocks are released from rest.After 2.5 s,the suspended block reaches the floor.What is the coefficient of kinetic friction between the table and the sliding block?

Accepted Answer

The answer of In the figure,two wooden blocks each of...

Question 7

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle.The coefficients of static and kinetic friction between the hockey puck and the metal ramp are  and  ,respectively.The puck's initial speed is  .What vertical height does the puck reach above its starting point?

Accepted Answer

The answer of A 200 g hockey puck is launched...

Question 8

A box is sliding down an incline tilted at a 12.0° angle above horizontal.The box is initially sliding down the incline at a speed of 1.50 m/s.The coefficient of kinetic friction between the box and the incline is 0.340.How far does the box slide down the incline before coming to rest?

Accepted Answer

The answer of A box is sliding down an incline...

Question 9

Two boxes are connected by a weightless cord running over a very light frictionless pulley as shown in the figure.Box A,of mass 8.0 kg,is initially at rest on the top of the table.The coefficient of kinetic friction between box A and the table is 0.10.Box B has a mass of 15.0 kg,and the system begins to move just after it is released.  (a)Draw the free-body diagrams for each of the boxes,identifying all of the forces acting on each one.
(b)Calculate the acceleration of each box.
(c)What is the tension in the cord?

Accepted Answer

The answer of Two boxes are connected by a weightless...

Question 10

A pickup truck is moving at 25 m/s with a toolbox of mass m resting on the bed of the truck 2.5 m behind the cab.Suddenly the brakes are applied,causing the toolbox to slide,and the truck comes to a stop in 4.7 s.The coefficient of kinetic friction between the toolbox and the bed of the truck is 0.28.After the brakes are applied,how much time elapses before the toolbox strikes the cab?

Accepted Answer

The answer of A pickup truck is moving at 25...

Question 11

A 4.00-kg box sits atop a 10.0-kg box on a horizontal table.The coefficient of kinetic friction between the two boxes and between the lower box and the table is 0.600,while the coefficient of static friction between these same surfaces is 0.800.A horizontal pull to the right is exerted on the lower box,as shown in the figure,and the boxes move together.What is the friction force on the UPPER box?

Accepted Answer

The answer of A 4.00-kg box sits atop a 10.0-kg...

Question 12

A packing crate rests on a horizontal surface.It is acted on by three horizontal forces: 600 N to the left,200 N to the right,and friction.The weight of the crate is 400 N.If the 600-N force is removed,the resultant force acting on the crate is

Accepted Answer

The crate is initially in equilibrium, since the forces to the right and left balance each other out, and the force of friction is equal and opposite to the resultant force. When the 600 N force to the left is removed, the forces to the right and left are no longer balanced, but the force of friction is still equal and opposite to the resultant force. Therefore, the net force on the crate is zero, and it remains in equilibrium.

Question 13

You push downward on a box at an angle 25° below the horizontal with a force of  .If the box is on a flat horizontal surface for which the coefficient of static friction with the box is 0.76,what is the mass of the heaviest box you will be able to move?

Accepted Answer

The answer of You push downward on a box at...

Question 14

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle.The coefficients of static and kinetic friction between the hockey puck and the metal ramp are  and  ,respectively.The puck's initial speed is 4.9 m/s.What speed does it have when it slides back down to its starting point?

Accepted Answer

The answer of A 200 g hockey puck is launched...

Question 15

A 4.00-kg block rests on a 30.0° incline as shown in the figure.If the coefficient of static friction between the block and the incline is 0.700,what magnitude horizontal force F must act on the block to start it moving up the incline?

Accepted Answer

The answer of A 4.00-kg block rests on a 30.0°...

Question 16

In the figure,a T-bar ski tow pulls a skier up a hill inclined at 10° above horizontal.The skier starts from rest and is pulled by a cable that exerts a tension T at an angle of 30° above the surface of the hill.The mass of the skier is 60 kg and the effective coefficient of kinetic friction between the skis and the snow is 0.100.What is the maximum tension in the cable if the starting acceleration is not to exceed 0.400 g?

Accepted Answer

The answer of In the figure,a T-bar ski tow pulls...

Question 17

A system comprised blocks,a light frictionless pulley,and connecting ropes is shown in the figure.The 9.0-kg block is on a perfectly smooth horizontal table.The surfaces of the 12-kg block are rough,with  between the block and the table.If the 5.0-kg block accelerates downward when it is released,find its acceleration.

Accepted Answer

The answer of A system comprised blocks,a light frictionless pulley,and...

Question 18

A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N which is parallel to the slope.The coefficient of kinetic friction between the block and the slope is 0.200.What is the magnitude of the acceleration of the block?

Accepted Answer

The answer of A 50.0-kg block is being pulled up...

Question 19

A rope pulls on the lower block in the figure with a tension force of 20 N.The coefficient of kinetic friction between the lower block and the surface is 0.16.The coefficient of kinetic friction between the lower block and the upper block is also 0.16.The pulley has no appreciable mass or friction.What is the acceleration of the 2.0 kg block?

Accepted Answer

First, we need to find the frictional force acting on the lower block. 
Frictional force = coefficient of kinetic friction * normal force 
Using the free-body diagram of the lower block, we can see that:
Normal force = Weight of lower block - Tension force = mg - 20, where g is the acceleration due to gravity (9.8 m/s²) and m is the mass of the lower block (2.0 kg).
Therefore, Frictional force = 0.16 * (mg - 20)
Next, we need to use the free-body diagram of the upper block to find the net force acting on it. We can see that:
Net force = Tension force - Frictional force
Using Newton's second law (F = ma), we can write:
(Weight of upper block - Tension force) - (0.16 * Tension force) = (mass of upper block) * (acceleration of blocks)
Simplifying and substituting given values, we get:
(5g - 20) - (0.16 * 20) = 2a
5g - 32 = 2a
a = (5g - 32) / 2
a ≈ 4.1 m/s²

Question 20

Block A of mass 5.0 kg and block X are attached to a rope which passes over a pulley,as shown in the figure.An 80-N force P is applied horizontally to block A,keeping it in contact with a rough vertical face.The coefficients of static and kinetic friction between the wall and block A are μ_s = 0.40 and μ_k = 0.30.The pulley is light and frictionless.The mass of block X is adjusted until block A moves upward with an acceleration of 1.6 m/s².What is the mass of block X?

Accepted Answer

First, we need to calculate the force of friction acting on block A. Since it is in contact with the rough vertical face, the frictional force will be equal to μ_s times the normal force. The normal force will be equal to the weight of block A, which is 5.0 kg times 9.8 m/s² or 49 N. Therefore, the frictional force is 0.40 times 49 N or 19.6 N. Next, we need to calculate the net force acting on block A. Since it is being pulled to the right with an 80-N force and the friction is acting to the left with a force of 19.6 N, the net force is (80 N - 19.6 N) or 60.4 N. The tension in the rope is the same on both sides of the pulley, so the force pulling block X upward is also 60.4 N. We can use this information to calculate the mass of block X: F = ma 60.4 N = (5.0 kg + m) × 1.6 m/s² m = 9.9 kg

Quiz 5: Applying Newtons Laws