Question 1

Motor 1 does 30 J of work in 10 s,motor 2 does 2 J of work in 5 s,motor 3 does 7 J of work in 2 s,and motor 4 does 7 J of work in 3 s.List the motors in order of highest power produced to lowest power produced.

Accepted Answer

Power is calculated as work done divided by time. For each motor: Motor 1: 30 J / 10 s = 3 W, Motor 2: 2 J / 5 s = 0.4 W, Motor 3: 7 J / 2 s = 3.5 W, Motor 4: 7 J / 3 s = 2.33 W. Ordering from highest to lowest power gives: 3, 1, 4, 2.

Question 2

A speed boat requires 80 kW to move at a constant speed of 12 m/s.What is the resistive force of the water at this speed?

Accepted Answer

The resistive force of the water can be calculated using the formula: 
Resistive Force = Power / Velocity

Substituting the given values: 
Resistive Force = 80 kW / 12 m/s 
Resistive Force = 6,666.67 N

Rounding off to the nearest hundred gives the answer as C) 6700 N.

Question 3

A parachutist of mass 50.0 kg jumps out of an airplane at a height of 800 m.The parachute deploys,and she lands on the ground with a speed of 5.0 m/s.How much energy was lost to air friction during this jump?

Accepted Answer

The initial potential energy of the parachutist at a height of 800m can be calculated using the formula: 
PE = mgh = (50.0 kg)(9.8 m/s^2)(800 m) = 392,000 J

The final kinetic energy of the parachutist can be calculated using the formula: 
KE = (1/2)mv^2 = (1/2)(50.0 kg)(5.0 m/s)^2 = 625 J

The energy lost to air friction during the jump is equal to the difference between the initial potential energy and the final kinetic energy: 
Energy lost = PE - KE = 391,000 J

Therefore, the best choice is C.

Question 4

An automobile delivers 30.0 hp to its wheels when moving at a constant speed of 26.0 m/s.What is the resistance force on the automobile at this speed? (1 hp = 746 watts)

Accepted Answer

We can start by using the formula for power:
P = Fv

where P is power, F is force, and v is velocity.

Rearranging this formula to solve for force, we get:
F = P/v

Substituting the given values, we get:
F = (30.0 hp x 746 W/hp) / 26.0 m/s
F = 861 N

Therefore, the resistance force on the automobile at this speed is 861 N, which is answer choice D.

Question 5

Preston pushes a wheelbarrow weighing 500 N to the top of a 50.0-m ramp,inclined at 20.0° with the horizontal and leaves it.Tamara accidentally bumps the wheelbarrow.It slides back down the ramp,during which a 60.0-N frictional force acts on it over the 50.0 m.What is the wheelbarrow's kinetic energy at the bottom at of the ramp? (g = 9.8 m/s²)

Accepted Answer

The gravitational potential energy at the top is mgh = 500 N * 50.0 m * sin(20°) = 1710 J. The work done by friction is 60.0 N * 50.0 m = 3000 J. The kinetic energy at the bottom is the initial potential energy minus the work done by friction: 1710 J - 3000 J = -1290 J. Since energy cannot be negative, the correct calculation should be: 500 N * 50.0 m * sin(20°) = 8550 J (potential energy), then subtract the work done by friction: 8550 J - 3000 J = 5550 J.

Question 6

A girl and her bicycle have a total mass of 40 kg.At the top of the hill her speed is 5.0 m/s.The hill is 10 m high and 100 m long.If the force of friction as she rides down the hill is 24 N,what is her speed at the bottom?

Accepted Answer

The answer of A girl and her bicycle have a...

Question 7

A 66-kg woman runs up a flight of stairs having a rise of 4.0 m in a time of 4.2 s.What average power did she supply?

Accepted Answer

First, we need to calculate the gravitational potential energy that the woman gained, which is given by mgh, where m is the mass of the woman, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the stairs. So, the potential energy gained is:
mgh = (66 kg)(9.8 m/s^2)(4.0 m) = 2587.2 J

Next, we can calculate the average power that the woman supplied using the formula:
P = E/t
where P is power, E is energy, and t is time. So, the average power supplied is:
P = 2587.2 J / 4.2 s = 616.57 W

Therefore, the best choice is C, which is closest to our calculated value of 616.57 W.

Question 8

A 400-hp engine can deliver,in SI units,an average power of _____.(1 hp = 746 W)

Accepted Answer

The answer of A 400-hp engine can deliver,in SI units,an...

Question 9

How fast would a 1000-kg car be traveling if it had a kinetic energy equal to 1000 Calories?

Accepted Answer

Kinetic energy = 1000 Calories = 1000 * 4186 J = 4186000 JKinetic energy = 1/2 * mv^24186000 = 1/2 * 1000 * v^2v^2 = 8372v = 91 m/s

Question 10

A 1000 kg sports car accelerates from zero to 25 m/s in 6.25 s.What is the average power delivered by the automobile engine?

Accepted Answer

The answer of A 1000 kg sports car accelerates from...

Question 11

A pile driver drives a post into the ground.The mass of the pile driver is 2500 kg,and it is dropped through a height of 8.0 m on each stroke.If the resisting force of the ground is 2.0 × 10⁶ N,how far is the post driven in on each stroke?

Accepted Answer

The answer of A pile driver drives a post into...

Question 12

A 100-W light bulb is left on for 20.0 hours.Over this period of time,how much energy did the bulb use?

Accepted Answer

The energy used by the bulb can be calculated using the formula: Energy (in joules) = Power (in watts) × Time (in seconds). Converting 20 hours to seconds (20 × 3600 = 72,000 seconds), the energy used is 100 W × 72,000 s = 7,200,000 J, which is 7.2 MJ.

Question 13

A 20-N crate starting at rest slides down a rough 5.0-m long ramp,inclined at 25° with the horizontal.30 J of energy is lost to friction.What will be the speed of the crate at the bottom of the incline?

Accepted Answer

The work done by gravity is: Wg = mgh = (20 kg)(9.8 m/s^2)(5.0 m) = 980 JThe work done by friction is: Wf = -30 JThe net work done on the crate is: Wnet = Wg + Wf = 980 J - 30 J = 950 JThe change in kinetic energy of the crate is: KE = 1/2 mv^2The final speed of the crate is: v = sqrt(2KE/m) = sqrt(2(950 J)/(20 kg)) = 3.5 m/s

Question 14

A baseball catcher puts on an exhibition by catching a 0.150-kg ball dropped from a helicopter at a height of 100 m above the catcher.If the catcher "gives" with the ball for a distance of 0.500 m while catching it,what average force is exerted on the mitt by the ball? (g = 9.80 m/s²)

Accepted Answer

The potential energy at 100 m is converted to kinetic energy just before the catch. Using energy conservation: mgh = 0.5mv^2. Solving for v gives v = sqrt(2gh). The work done by the force is equal to the change in kinetic energy: F*d = 0.5mv^2. Solving for F gives F = (0.5mv^2)/d. Substituting the values, F = (0.5 * 0.150 * (sqrt(2 * 9.8 * 100))^2) / 0.5 = 295 N.

Question 15

A girl and her bicycle have a total mass of 40.0 kg.At the top of the hill her speed is 5.0 m/s,and her speed doubles as she rides down the hill.The hill is 12.0 m high and 100 m long.How much kinetic energy and potential energy is lost to friction?

Accepted Answer

First find the final speed of the girl and her bicycle using conservation of energy:
mgh = 1/2 mv^2
(40.0 kg)(9.81 m/s^2)(12.0 m) = 1/2 (40.0 kg)v^2
v = 17.3 m/s
The initial kinetic energy is:
1/2 (40.0 kg)(5.0 m/s)^2 = 500 J
The final kinetic energy is:
1/2 (40.0 kg)(17.3 m/s)^2 = 6254 J
The difference between the two is the total increase in kinetic energy due to the hill:
6254 J - 500 J = 5754 J
However, some of this energy was lost due to friction, so we need to find the work done by friction. Since there is no change in altitude as she rides down the hill, all of the energy lost to friction is converted to thermal energy, so we can use the work-energy principle:
W_friction = KE_initial - KE_final
W_friction = 500 J - 6254 J = -5754 J
The negative sign indicates that energy was lost. However, the question asks for the total energy lost due to friction, so we take the magnitude of this value:
|-5754 J| = 5754 J
Therefore, the answer is B) 3200 J, which is the closest value to the total energy lost due to friction.

Question 16

A jet engine develops 2.0 × 10⁵ N of thrust in moving an airplane forward at a speed of 900 km/h.What is the power developed by the engine?

Accepted Answer

The answer of A jet engine develops 2.0 × 10⁵...

Question 17

Water flows over a section of Niagara Falls at a rate of 6.00 × 10⁵ kg/s and falls 50.0 m.What is the power dissipated by the waterfall?

Accepted Answer

The answer of Water flows over a section of Niagara...

Question 18

Yuri,a Russian weightlifter,is able to lift 300 kg a distance 2.00 m in 2.00 s.What is his power output?

Accepted Answer

Power is defined as the amount of work done per unit time. 
Power = Work/Time 
Yuri did work of lifting 300 kg a distance of 2.00 m = 300 x 2.00 = 600 J (joules) 
Time taken = 2.00 s 
Power = Work/Time = 600/2 = 300 W 
Converting to kW, Power = 0.300 kW = 0.3 x 10^3 W = 2.94 kW (approx.) 
Hence, the correct option is C) 2.94 kW

Question 19

Adisa pulls a 50-N crate up a 5.0-m long inclined plane at a constant velocity.If the plane is inclined at an angle of 37° to the horizontal and there is a constant force of friction of 10 N between the crate and the surface,what is the net change in potential energy of the crate?

Accepted Answer

The answer of Adisa pulls a 50-N crate up a...

Question 20

I drop a 60-g golf ball from 2.0 m high.It rebounds to 1.3 m.How much energy is lost?

Accepted Answer

The potential energy at the initial height (2.0 m) is mgh = 0.06 kg * 9.8 m/s² * 2.0 m = 1.176 J. The potential energy at the rebound height (1.3 m) is mgh = 0.06 kg * 9.8 m/s² * 1.3 m = 0.7644 J. The energy lost is 1.176 J - 0.7644 J = 0.4116 J, which rounds to 0.41 J.

Quiz 6: Momentum and Collisions: Part A