True/false Freshman Undergrad Physics Problems (Work)

[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

 

[NBAJam100]

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

 

[TwinGemini14]

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

ANSWER = C

 

[TwinGemini14]

Are these right? If not, can somebody help?

 

[TwinGemini14]

True/false Physics Problems about Work

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

ANSWER = C

----
Can anybody help me with these?

 

[LowlyPion]

Each cart has the same kinetic energy.

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

 

[LowlyPion]

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

 

[LowlyPion]

3 looks correct.

 

[Kyriam]

DISCLAIMER: These are just my opinions, I have been wrong often enough.

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.

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.

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

ANSWER = C

I agree.

 

[spooshah]

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.