a horizontal force of magnitude 35.0N pushes a block of mass 4.0kg across a floor where the coefficient of?

Question

a horizontal force of magnitude 35.0N pushes a block of mass 4.0kg across a floor where the coefficient of kinetic friction is 0.600. 1)how much work is done by that applied force on the block - floor system when the block slides through a displacement of 3.00m across the floor? 2)during that displacement, the thermal energy of the block increases by 40.0J. what is the increase in thermal energy of the floor? 3)what is the increase in kinetic energy of the block?

full solution pleas.

Answer 1

The most important concept in this question is to check what is the friction force acting actually. For that calculate the maximum friction force that can act.
coefficient of friction * mass of block * acc. due to gravity
= 0.6 * 4 * 9.81
= 23.544 N
Now the applied force is 35 N which is more than the maximum friction force. This means that maximum friction force will act.
Work Done by applied force = Force * Displacement
= 35*3 = 105 J
Work Done by friction force = Friction force * Displacement
= 23.544*3 = 70.632 J
Increase in thermal energy of floor = 70.632 - 40 = 30.632 J
This is because work done by friction goes into heat only.
Work that has been utilized in accelerating the block = 105 - 70.632 = 34.368 J

Answer 2

be conscious that an same volume of pressure is utilized on pushing both m? and m? in the course of the floor, so we see that the relationship between m? and m? is: (pressure pushing block of mass m?) = (pressure pushing block of mass m?) ==> m?(10 m/s^2) = m?(3 m/s^2), considering that F = ma ==> m? = (3/10)m?. (a) A block of mass m? - m? has a mass of: m? - m? = m? - (3/10)m? = (a million - 3/10)m? = (7/10)m?, so considering that F = ma implies a = F/m, the acceleration of the block is inversely proportional to its mass. because the block of mass m? - m? has 7/10 situations the mass of a block with mass m?, the acceleration will be a million/(7/10) = 10/7 situations as a lot. subsequently, the necessary acceleration is: (3 m/s^2)(10/7) = 30/7 m/s^2 = 4.3 m/s^2, to 2 s.f. . (b) A block of mass m? + m? has mass: m? + m? = m? + (3/10)m? = (a million + 3/10)m? = (13/10)m?, so from the communicate partly (a), the block of m? + m? has a million/(13/10) = 10/13 situations the acceleration of the block of mass m?; subsequently, the acceleration of the block of mass m? + m? is: (3 m/s^2)(10/13) = 30/13 m/s^2 = 2.3 m/s^2, to 2.s.f. . i'm hoping this helps!

Answer 3

Not a full solution; if you can't do arithmetic you're taking the wrong course!
The total work done, answer 1, is F*D (35*3).
The friction force is m*g*friction coefficient. Multiply that by 3 m for the frictional work done. Then subtract the block's thermal energy increase from the friction work for answer 2.
Subtract the frictional work from the total work to get the work done accelerating the block. That's it's kinetic energy, answer 3.
EDIT: The next answer follows my formulas and does all the arithmetic for you, saving you the trouble of using a calculator.