Question 1

A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to 15.9 m/s.  How far does it travel while slowing down?

Accepted Answer

The work done by the net force on the car is equal to the change in kinetic energy of the car. The work done by the net force is given by:$$W = Fd$$where:* W is the work done (in joules)* F is the net force (in newtons)* d is the distance traveled (in meters)The change in kinetic energy of the car is given by:$$KE = \frac{1}{2}mv^2$$where:* KE is the kinetic energy (in joules)* m is the mass of the car (in kilograms)* v is the velocity of the car (in meters per second)Setting the work done by the net force equal to the change in kinetic energy, we get:$$Fd = \frac{1}{2}mv^2_f - \frac{1}{2}mv^2_i$$where:* v_f is the final velocity of the car (in meters per second)* v_i is the initial velocity of the car (in meters per second)Substituting the given values into the equation, we get:$$9500 d = \frac{1}{2}(1000)(15.9)^2 - \frac{1}{2}(1000)(30)^2$$$$9500 d = -115,550$$$$d = -12.16 m$$The negative sign indicates that the car is traveling in the opposite direction of the net force. Therefore, the distance traveled by the car while slowing down is 12.16 m.

Question 2

A sled having a certain initial speed on a horizontal surface comes to rest after traveling 10 m.If the coefficient of kinetic friction between the object and the surface is 0.20,what was the initial speed of the object?

Accepted Answer

The initial speed of the object can be found using the work-energy principle. The work done by friction is equal to the change in kinetic energy. The work done by friction is $W = f_k \cdot d = \mu_k \cdot m \cdot g \cdot d$, where $\mu_k$ is the coefficient of kinetic friction, $m$ is the mass of the object, $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$), and $d$ is the distance (10 m). The change in kinetic energy is $\Delta KE = \frac{1}{2} m v^2 - 0$, since the object comes to rest. Setting the work done by friction equal to the change in kinetic energy and solving for $v$ gives $v = \sqrt{2 \cdot \mu_k \cdot g \cdot d}$. Substituting $\mu_k = 0.20$, $g = 9.8 \, \text{m/s}^2$, and $d = 10 \, \text{m}$, we find $v \approx 6.3 \, \text{m/s}$.

Question 3

A driver,traveling at 22 m/s,slows down her 2000 kg truck to stop for a red light.What work is done on the truck by the friction force of the road?

Accepted Answer

The answer of A driver,traveling at 22 m/s,slows down her...

Question 4

An ideal spring stretches by 21.0 cm when a 135-N object is hung from it.If instead you hang a fish from this spring,what is the weight of a fish that would stretch the spring by

Accepted Answer

The answer of An ideal spring stretches by 21.0 cm...

Question 5

A 10-kg mass,hung by an ideal spring,causes the spring to stretch 2.0 cm.What is the spring constant (force constant)for this spring?

Accepted Answer

The spring constant (k) can be calculated using the formula k=F/x, where F is the force applied and x is the displacement. In this case, the force applied is the weight of the mass, which is F = m*g = 10*9.8 = 98 N (taking g as 9.8 m/s^2). The displacement is given as x = 2.0 cm = 0.02 m. Therefore, k = 98/0.02 = 4900 N/m = 49 N/cm. The best choice is B.

Question 6

How large a net force is required to accelerate a 1600-kg SUV from rest to a speed of 25 m/s in a distance of 200 m?

Accepted Answer

We can use the formula:

distance = (1/2) * acceleration * time^2

to solve for the acceleration required to cover a distance of 200 m with an initial velocity of 0 m/s:

200 = (1/2) * a * t^2

Since the SUV starts from rest, its initial velocity is 0 m/s, so we can also use the formula:

final velocity^2 = initial velocity^2 + 2 * acceleration * distance

to solve for the acceleration required to reach a final velocity of 25 m/s after covering a distance of 200 m:

25^2 = 0^2 + 2 * a * 200

Simplifying, we find that:

a = 6.25 m/s^2

Finally, we can use Newton's second law of motion (F = m * a) to determine the net force required:

F = 1600 kg * 6.25 m/s^2 = 10,000 N ≈ 2500 N

Therefore, the answer is E) 2500 N.

Question 7

An ideal spring has a spring constant (force constant)of 2500 N/m.is stretched 4.0 cm.How much work is required to stretch the spring by 4.0 cm?

Accepted Answer

The work required to stretch or compress a spring is given by the formula $ W = \frac{1}{2} k x^2 $, where $W$ is the work done, $k$ is the spring constant, and $x$ is the displacement from the equilibrium position in meters. Given $k = 2500 \, \text{N/m}$ and $x = 4.0 \, \text{cm} = 0.04 \, \text{m}$, the work done is $ W = \frac{1}{2} \times 2500 \, \text{N/m} \times (0.04 \, \text{m})^2 = 2.0 \, \text{J}$.

Question 8

A certain car traveling at 34.0 mph skids to a stop in 29 meters from the point where the brakes were applied.In approximately what distance would the car have stopped had it been going 105.4 mph?

Accepted Answer

The stopping distance is proportional to the square of the velocity. If the velocity triples (from 34.0 mph to 102 mph, approximately 3 times), the stopping distance would increase by a factor of 3^2 = 9. Since the original stopping distance is 29 meters, the new stopping distance would be 29 meters * 9 = 261 meters. Option A, 279 meters, is the closest to this calculated value, acknowledging some approximation in the velocity increase from the exact 105.4 mph given.

Question 9

A 100-N force has a horizontal component of 80 N and a vertical component of 60 N.The force is applied to a cart on a level frictionless floor.The cart starts from rest and moves 2.0 m horizontally along the floor due to this force.What is the cart's final kinetic energy?

Accepted Answer

The work done by a force is given by the formula W = Fd cos θ, where F is the force, d is the displacement, and θ is the angle between the force and the displacement.

In this case, the horizontal component of the force is causing the displacement, so θ = 0, and cos θ = 1. Therefore, the work done by the force is:

W = 80 N × 2.0 m × 1 = 160 J

Since there is no friction, all the work done by the force is converted into kinetic energy. Therefore, the cart's final kinetic energy is also 160 J. Hence, the correct answer is B.

Question 10

In a ballistics test,a 28-g bullet pierces a sand bag that is 
Thick.If the initial bullet velocity was 
And it emerged from the sandbag moving at 
What was the magnitude of the friction force (assuming it to be constant)that the bullet experienced while it traveled through the bag?

Accepted Answer

The answer of In a ballistics test,a 28-g bullet pierces...

Question 11

A child pulls on a wagon with a force of 
If the wagon moves a total of 
In 
What is the average power delivered by the child?

Accepted Answer

The answer of A child pulls on a wagon with...

Question 12

A stone is moving on a rough level surface.It has 24 J of kinetic energy,and the friction force on it is a constant 0.50 N.What is the maximum distance it can slide?

Accepted Answer

The answer of A stone is moving on a rough...

Question 13

On an alien planet,an object moving at 4.0 m/s on the horizontal ground comes to rest after traveling a distance of 10 m.If the coefficient of kinetic friction between the object and the surface is 0.20,what is the value of g on that planet?

Accepted Answer

Explanation:The work done by friction is:$$W_f=F_fd=\mu mgd$$The change in kinetic energy is:$$K_i-K_f=\frac{1}{2}mv_i^2-0=\frac{1}{2}mv_i^2$$By the work-energy theorem, we have:$$W_f=K_i-K_f$$$$\mu mgd=\frac{1}{2}mv_i^2$$$$g=\frac{v_i^2}{2\mu d}=\frac{(4.0	ext{ m/s})^2}{2(0.20)(10	ext{ m})}=\boxed{4.0	ext{ m/s}^2}$$

Question 14

A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m.What is the coefficient of kinetic friction between the stone and the surface?

Accepted Answer

The answer of A stone initially moving at 8.0 m/s...

Question 15

The kinetic friction force that a horizontal surface exerts on a 60.0-kg object is 50.0 N.If the initial speed of the object is 25.0 m/s,what distance will it slide before coming to a stop?

Accepted Answer

We can use the kinetic energy equation to solve for the distance:
KE = 1/2 mv^2
When the object comes to a stop, all the kinetic energy is converted to work done by friction:
KE = work = force x distance
Therefore, we can set these two equations equal to each other:
1/2 mv^2 = Fd
Solving for d, we get:
d = (1/2 mv^2)/F
Plugging in the given values, we get:
d = (1/2 x 60.0 kg x (25.0 m/s)^2)/50.0 N
d = 375 m
Therefore, the answer is C) 375 m.

Quiz 7: Work and Kinetic Energy

A 1000-kg car experiences a net force of 9500 N while slowing down from 30 m/s to 15.9 m/s. How far does it travel while slowing down?