What is the acceleration of a block being pulled up a slope by a 140 N force?

[joepro]

Homework Statement

Calculate the acceleration for a 10.0kg block which is being pulled up a 35 degree slope by a force of 140 N if the coefficient of friction is 0.100 The answer is to be in m/s2

Homework Equations

Ff=UN

The Attempt at a Solution

I am having trouble getting started and putting the parts of the problem together.

 

[ehild]

Start with a picture and free-body diagram.

ehild

 

[joepro]

I have a free body which a 140 N force going along the + x-axis a N force (normal) going + along Y axis a vector "W" in the third quad with a polar angle of 235'. What do i do next. I know that i need to find the acceleration of the block. Where and how does the coeficient of friction come into play here?

 

[ehild]

I still need that picture with all forces shown. There is gravity and friction in addition to that 140 N force. The resultant force accelerates the block.

ehild

 

[rwisz]

To solve this problem, we need to use Newton's Second Law, which states that the net force acting on an object equals its mass multiplied by its acceleration (F=ma). In this case, the net force is the force pulling the block up the slope (140 N) minus the force of friction (Ff=UN). We also need to consider the angle of the slope, as this will affect the components of the forces acting on the block.

First, we need to find the component of the force pulling the block up the slope in the direction of motion. This can be calculated by multiplying the force (140 N) by the sine of the angle of the slope (35 degrees). This gives us a force of approximately 80.5 N.

Next, we can calculate the force of friction by multiplying the coefficient of friction (0.100) by the normal force, which is the force acting perpendicular to the surface of the slope. The normal force can be calculated by multiplying the mass of the block (10.0 kg) by the acceleration due to gravity (9.8 m/s^2) and the cosine of the angle of the slope (35 degrees). This gives us a normal force of approximately 85.9 N. Therefore, the force of friction is 0.100 x 85.9 N, which is approximately 8.6 N.

Now, we can plug these values into our equation for net force (F=ma). The net force is 80.5 N (force pulling the block up the slope) minus 8.6 N (force of friction). This gives us a net force of approximately 71.9 N. We can then rearrange the equation to solve for acceleration (a=F/m). Plugging in the net force and the mass of the block, we get an acceleration of approximately 7.2 m/s^2.

In conclusion, the acceleration of the block being pulled up the slope by a 140 N force is 7.2 m/s^2.