a 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s, and at point B it has slowed to 1.25m/s.
1.) How much work was done on the book between A and B?
2.) If -0.750 J of work is done on the book from B to C, how fast is it moving at point C?
3.) How fast would it be moving at C if +0.750 J of work were done on it from B to C?
4.) The book slides a distance of 2.50m from A to B. Determine the coefficient from A to B whilst using constant-force work formula to recalculate the answer to #1.)
I'm having real trouble with this problem.
a jogger run around a square city block with sides of 30m. if she start an end at point a what distance has she run
To solve this problem, we will use the work-energy theorem and the concept of work done by a force. I'll break down each question separately to help you understand the process.
1.) To calculate the work done on an object, you need to know the change in its kinetic energy between two points. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
The formula for work done is given by:
Work = Change in Kinetic Energy = (1/2) * mass * (final velocity^2 - initial velocity^2)
In this case, we need to find the change in kinetic energy between points A and B. The mass of the book is 1.50 kg, and the initial velocity at point A is 3.21 m/s, while at point B it is 1.25 m/s.
Using the formula, we can calculate the work done:
Work = (1/2) * 1.50 kg * ((1.25 m/s)^2 - (3.21 m/s)^2)
2.) To find the final velocity at point C, given that -0.750 J of work is done on the book from B to C, we can apply the work-energy theorem again. The work done is negative, indicating a loss of kinetic energy.
Since the work done is equal to the change in kinetic energy, we can calculate the change in kinetic energy. The final kinetic energy is less than the initial kinetic energy, so the magnitude of the change in kinetic energy will be |-0.750 J|.
Using the formula for work done:
|-0.750 J| = (1/2) * 1.50 kg * (v_c^2 - 1.25 m/s)^2)
Solve for v_c (the velocity at point C).
3.) The same formula can be used to find the velocity at point C if +0.750 J of work is done on the book from B to C. In this case, the work done is positive, indicating a gain in kinetic energy. You will use the same formula as question 2, but with a positive value for the change in kinetic energy.
4.) To find the coefficient of friction between points A and B, we can use the constant-force work formula:
Work = Force * Distance * cos(theta)
In this case, the work done is already calculated as given in question 1. The distance is given as 2.50 m. By rearranging the equation, we can solve for the force:
Force = Work / (Distance * cos(theta))
Since the gravitational force (mg) and the normal force (mgcos(theta)) cancel each other out in the horizontal direction, the only force acting on the book is the frictional force.
The coefficient of friction (μ) can be calculated using the formula:
μ = Frictional Force / Normal Force
The normal force is equal to the weight of the book (mg). Substitute the calculated frictional force and normal force into the equation to find the coefficient of friction.
Horizontal surface, no change in PE.
a) work done is by friction...what is the difference in KE
b) same principle
c) same principle
4) friction=work/distance
mu*mg=work/distance solve for mu