Block slides down a slope question (force and time)

[physicsguy112]

Homework Statement

A small block slides down a slanted board when released. The upper half of the board is smooth and the lower is rough, so that the acceleration of the block on the smooth half is three times greater than it is on the rough half. The block reaches the bottom of the board in time t1. The board is then flipped so that the upper half is rough and the lower part is smooth, and the block is released from the top again. This time, the block reaches the bottom of the board in time t2. In both cases, the board makes the same angle with the horizontal. Find the ratio t1/t2.

Homework Equations

Vf=Vo+at

F=ma

The Attempt at a Solution

I think I'm going to have to solve it in 2 parts, solve for t1 and t2 separately. When solving for t1 or t2, do I need to split up the rough and smooth surfaces?

I'm having trouble getting this one setup. I would really appreciate a strategy to help me solve this one. Thanks in advance for the help.

 

[Nate810]

For t1:We can set up the equations for the two parts of the board. For the smooth half: Vf = Vo + at 0 = Vo + 3at Vo = -3at For the rough half: Vf = Vo + at Vf = -3at + at Vf = -2at Now, we can combine the equations to get: 0=-3at -2at -5at = 0 at = 0 So, the time it takes for the block to reach the bottom of the board is: t1 = Vo/a = 0/0 = undefined For t2: We can set up the equations for the two parts of the board. For the rough half: Vf = Vo + at 0 = Vo + at Vo = -at For the smooth half: Vf = Vo + at Vf = -at + 3at Vf = 2at Now, we can combine the equations to get: 0=-at -2at -3at = 0 at = 0 So, the time it takes for the block to reach the bottom of the board is: t2 = Vo/a = 0/0 = undefined Therefore, the ratio of t1/t2 is undefined.

 

[kerimek]

I would approach this problem by first identifying the key variables and principles involved. In this case, the variables are the acceleration of the block on the smooth and rough surfaces, the time it takes for the block to reach the bottom of the board, and the angle of the board with the horizontal. The principle involved is Newton's second law of motion, which states that the force acting on an object is equal to its mass times its acceleration (F=ma).

Next, I would consider the forces acting on the block as it slides down the board. On the smooth surface, the only force acting on the block is its weight (mg), which causes it to accelerate downward. On the rough surface, there is an additional force of friction acting on the block, which slows down its acceleration. This can be expressed as Ff=μN, where μ is the coefficient of friction and N is the normal force.

Using these principles, I would set up equations for the acceleration of the block on the smooth and rough surfaces, and then solve for t1 and t2 separately. For example, for the smooth surface, we can use the equation Vf=Vo+at and solve for t1:

t1 = (Vf-Vo)/a

Since we know that the acceleration on the smooth surface is three times greater than on the rough surface, we can set up a ratio:

a smooth = 3a rough

Substituting this into our equation for t1, we get:

t1 = (Vf-Vo)/(3a rough)

We can do a similar process for the rough surface, using the equation F=ma to solve for t2. Once we have solved for t1 and t2 separately, we can find the ratio t1/t2.

In summary, the key to solving this problem is to identify the relevant variables and principles involved, and then set up equations based on those variables and principles. By doing this separately for the smooth and rough surfaces, we can find the ratio t1/t2 and answer the question.