# True/false Freshman Undergrad Physics Problems (Work)

• TwinGemini14
In summary: If so, the final potential energy is the same, and the final kinetic energy is the same. The latter doesn't depend on mass.
TwinGemini14
1. Two carts, one of mass M and one of mass 2M, are sliding on a horizontal frictionless surface with the same initial kinetic energy. A constant force is applied to each cart in order to bring them to rest after the same distance. The less massive cart requires a larger force to come to rest.

True or False

2. The final kinetic energy of an object dropped from rest through a fixed vertical distance does not depend on its mass.

True or False

3. A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s.

How much work is done on the box by the rope?

-10 J
0 J
10 J

did you make any attempt at solving these problems on your own?

Yeah,

1. Two carts, one of mass M and one of mass 2M, are sliding on a horizontal frictionless surface with the same initial kinetic energy. A constant force is applied to each cart in order to bring them to rest after the same distance. The less massive cart requires a larger force to come to rest.

This one seemed really confusing to me...

Since they have the same KE,

i. 0.5*m*(sqrt(2)v)^2
ii. 0.5*(2m)*v^2

0.5*m*(sqrt(2)v)^2 = 0.5*(2m)*v^2

So... since there is a constant force, a=0
F = ma
F = m(0) = 0

So are they both the same force? := FALSE?

2. The final kinetic energy of an object dropped from rest through a fixed vertical distance does not depend on its mass.

Since 0.5mv^2 = mgh
Mass cancels out... So...

TRUE?

3. A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s.
How much work is done on the box by the rope?

Since it is asking for the work done by the rope and not the net force (friction subtracted), isn't it simply...

W = Fd = (5)(2) = 10J

Are these right? If not, can somebody help?

## Homework Statement

1. Two carts, one of mass M and one of mass 2M, are sliding on a horizontal frictionless surface with the same initial kinetic energy. A constant force is applied to each cart in order to bring them to rest after the same distance. The less massive cart requires a larger force to come to rest.

True / False

2. The final kinetic energy of an object dropped from rest through a fixed vertical distance does not depend on its mass.

True / false

3. A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s.
How much work is done on the box by the rope?

A. -10J
B. 0J
C. 10J

## Homework Equations

KE = 0.5mv^2
W = Fd
W = KE2 - KE1

3. The Attempt at a Solution

1. Two carts, one of mass M and one of mass 2M, are sliding on a horizontal frictionless surface with the same initial kinetic energy. A constant force is applied to each cart in order to bring them to rest after the same distance. The less massive cart requires a larger force to come to rest.

This one seemed really confusing to me...

Since they have the same KE,

i. 0.5*m*(sqrt(2)v)^2
ii. 0.5*(2m)*v^2

0.5*m*(sqrt(2)v)^2 = 0.5*(2m)*v^2

So... since there is a constant force, a=0
F = ma
F = m(0) = 0

So are they both the same force? := FALSE?

2. The final kinetic energy of an object dropped from rest through a fixed vertical distance does not depend on its mass.

Since 0.5mv^2 = mgh
Mass cancels out... So...

TRUE?

3. A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s.
How much work is done on the box by the rope?

Since it is asking for the work done by the rope and not the net force (friction subtracted), isn't it simply...

W = Fd = (5)(2) = 10J

----
Can anybody help me with these?

Each cart has the same kinetic energy.

Each is to be stopped over the same distance. That sounds like Work which is what?

For the dropped object ... what is the formula for Kinetic energy again?

3 looks correct.

DISCLAIMER: These are just my opinions, I have been wrong often enough.
TwinGemini14 said:
1. Two carts, one of mass M and one of mass 2M, are sliding on a horizontal frictionless surface with the same initial kinetic energy. A constant force is applied to each cart in order to bring them to rest after the same distance. The less massive cart requires a larger force to come to rest.

Since they have the same KE, ...
Since the gravitational potential energy does not change and there is no friction (horizontal frictionless surface), the change in total and kinetic energy is equal to the work of the constant braking force. The change in kinetic energy (from equal initial KE to zero) is the same for both carts, so the work by the braking force is the same. Since the distance is the same, the force must be the same.

TwinGemini14 said:
2. The final kinetic energy of an object dropped from rest through a fixed vertical distance does not depend on its mass.

Since 0.5mv^2 = mgh
Mass cancels out... So...
I would agree that the "final velocity" is the same.

Let's assume that there is no air friction. The acceleration of gravity is independent of the mass, and the initial velocity is zero. Two objects dropped at the same time will have the same acceleration, and velocity at all times. They also will have covered the same distance at all times, regardless of mass. As they both reach the same vertical distance, they will have the same "final velocity." I agree that far.

The kinetic energy (0.5mv^2) calculated based on the final velocity will vary depending on the mass, though.

TwinGemini14 said:
3. A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s.
How much work is done on the box by the rope?

A. -10J
B. 0J
C. 10J

## Homework Equations

KE = 0.5mv^2
W = Fd
W = KE2 - KE1

3. The Attempt at a Solution

Since it is asking for the work done by the rope and not the net force (friction subtracted), isn't it simply...

W = Fd = (5)(2) = 10J

I agree.

Hi ,
I might be too late inreplying but it may help in future understanding.
I will only attempt the 1st task as it is interesting and classical mechanics... The other 2 are simple enough to be understood...
You have already got the correct answer for task 1 but I am just attempting an explanation...

Suppose m1 = 4 * m2 ,
=> v2 = 2 * v1 (bcoz. m1*v1^2 = m2 * v2^2),

Neglecting friction suppose we apply the same force
the deceleration produced would be
a1 = F / m1 and a2 = F / m2 ,
=> a2 = 4 * a1 ( Bcoz m1 = 4 * m2 )

a = v / t => time required for a body to go from v to 0 is t = (v1 - 0) / a,

=> t1 = v1 / a1 and t2 = v2 / a2 ,
Thus t2 = t1 / 2 (Bcoz. a2 = 4 * a1)

Hence the time required for each body to stop will be different and depend upon the mass.

However , let us consider the distance travelled. For a constant deceleration the average velocity during the deceleration will be ( v - 0 {final velocity} ) / 2

Hence Avg_v1=v1 /2 and Avg_v2 = v2 /2,

=>Avg_v1 = 2 * Avg_v2 (Bcoz. v2 = 2 * v1),

Distance traveled is velocity * time
d1 = v1 /2 * t1 and d2 = v2 /2 *t2
=> d2 = v1 * t1 /2 (Bcoz. v2 = 2 * v1 and t2 = t1 /2).

Hence d1 = d2. But t1 /= t2.

## 1. What is the purpose of "True/false Freshman Undergrad Physics Problems (Work)"?

The purpose of these types of problems is to assess a freshman undergrad student's understanding and application of basic physics concepts related to work. These problems are designed to test a student's ability to identify whether a statement about a specific physics scenario is true or false.

## 2. What is the definition of work in physics?

In physics, work is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. This means that work is only considered to be done when a force causes a displacement in the same direction.

## 3. How is work calculated in physics?

The formula for calculating work in physics is W = F * d * cos(theta), where W is work, F is the applied force, d is the displacement, and theta is the angle between the force and displacement vectors. This formula can also be simplified to W = F * d if the force and displacement are in the same direction.

## 4. What are some examples of work in daily life?

Some examples of work in daily life include lifting a weight, pushing a cart, pulling a suitcase, and climbing stairs. Any activity that requires a force to cause a displacement can be considered as work in physics.

## 5. How can I improve my understanding of work in physics?

To improve your understanding of work in physics, it is important to practice solving problems and to review the basic concepts and formulas regularly. You can also seek help from a physics teacher or tutor if you are struggling to understand a specific concept related to work.

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