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Let's say you measure an average acceleration to be 0.3883 ms-2. What is the height (in centimeters) of the riser block? Assume...Asked by Richard
Let's say you measure an average acceleration to be 0.4134 ms-2.
What is the height (in centimeters) of the riser block?
Assume the legs of the track span 1.0000 m and that there is neither friction nor drag. Also assume the block is under one of the legs.
?? cm
Use three significant figures or N/A if not enough information is given.
I have seen this question posted before however I have not been able to determine how to solve it.
I know that some background information about the question is required:
• There is a glider track that creates an incline with a wooden block under one end of the track. This creates a right triangle where H is the height of the block and L (1.0000m) is the length of the hypotenuse.
• The acceleration refers to a small car that rolls down the incline (assuming no friction or drag acts on the car).
I have tried two things:
1.) (1/2)g(H/L) = 0.4134 solving for H this is incorrect. (I'm not sure if this is even close)
2.) I have attempted to solve with a free-body diagram:
Fx: m * g * Sin(theta)
Fnet = m * a
Fnet = m * g * Sin(H/L)
m * a = g * Sin(H/L)
a = g * Sin(H/L) Where I could solve for H. (I do not know if this is correct)
I can provide additional information if necessary, but I just need a direction to work in..
What is the height (in centimeters) of the riser block?
Assume the legs of the track span 1.0000 m and that there is neither friction nor drag. Also assume the block is under one of the legs.
?? cm
Use three significant figures or N/A if not enough information is given.
I have seen this question posted before however I have not been able to determine how to solve it.
I know that some background information about the question is required:
• There is a glider track that creates an incline with a wooden block under one end of the track. This creates a right triangle where H is the height of the block and L (1.0000m) is the length of the hypotenuse.
• The acceleration refers to a small car that rolls down the incline (assuming no friction or drag acts on the car).
I have tried two things:
1.) (1/2)g(H/L) = 0.4134 solving for H this is incorrect. (I'm not sure if this is even close)
2.) I have attempted to solve with a free-body diagram:
Fx: m * g * Sin(theta)
Fnet = m * a
Fnet = m * g * Sin(H/L)
m * a = g * Sin(H/L)
a = g * Sin(H/L) Where I could solve for H. (I do not know if this is correct)
I can provide additional information if necessary, but I just need a direction to work in..
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