What is the acceleration of the block?

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A 50.0-kg block is being pulled up a 16.0° slope with a force of 250 N, and the coefficient of kinetic friction is 0.200. The acceleration of the block is calculated to be 0.412 m/s², but the method to arrive at this answer is unclear to some participants. It is confirmed that the block experiences a normal force, as it is resting on the slope, despite confusion about whether tension is involved. Participants emphasize the importance of sketching a free body diagram to identify all forces acting on the block, including gravity, normal force, friction, and the applied force. Understanding these forces is crucial for solving the problem correctly.
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Homework Statement


A 50.0-kg block is being pulled up a 16.0° slope by a force of 250 N that is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the acceleration of the block?

Homework Equations


Fk=μkN
F=MA

The Attempt at a Solution


answer is .412 m/s2. I know the answer I just don't know the steps to solve this equation.
It is saying it is being pulled up, does that mean it has tension instead of normal force? it is not sitting on anything so it can not have a normal force right?
 
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Angela_vaal said:
answer is .412 m/s2. I know the answer I just don't know the steps to solve this equation.
You don't even have an equation to solve, and how do you know that is the right answer?
It is saying it is being pulled up, does that mean it has tension instead of normal force? it is not sitting on anything so it can not have a normal force right?
Surely the block is sittig on a slope? A tension is a force. Is the block being pulled by a rope or has someone just grabbed it and pulled it? Does it matter to the question?

Your first step is to sketch a free body diagram ... you will have course notes about this: so how many forces are on the block, and what are they?
 
Simon Bridge said:
Surely the block is suittig on a slope?
Your first step is to sketch a free body diagram ... you will have course notes about this: so how many forces are on the block, and what are they?

In that case, there is a normal force. So the forces on the block are the Normal force, MG and 250cos16, 250 sin16
 
Identify the forces themselves - you have a normal force to the slope, gravity, friction, and the pulling force.
GIve them names - so you have ##F_g=mg## pointing vertically down, ##N## points upwards at 90deg to the slope, ##f=\mu N## points down the slope and ##F=250##N points up the slope. F is only one force, not two like you wrote. Don't do the cosine and sine stuff untilyou decide how to orient your axes - hint: point x-axis up the slope.
 

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