Man running up a wheeled block.

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SUMMARY

The discussion focuses on a physics problem involving a man running up a wheeled block with mass M on a smooth surface. The key equations derived include the man's acceleration relative to the block, expressed as a = 2Lcos(α)/T², and the need to apply conservation of momentum to analyze the system. The acceleration of the man relative to the ground is also addressed, with the formula a = [2Lcos(α)/T² - Acos(α)]x + [N/m - gsin(α) - Asin(α)/m]y. The participants emphasize the importance of using momentum conservation rather than forces and gravity for this scenario.

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Hi,

Homework Statement


A man, whose mass is m, is running up a block of mass M (as shown in the diagram, please see attachment) with constant acceleration. The block is laid on a smooth ground. The man starts running from rest and the time it takes for him to traverse a distance L (across the block) is T.
(a) What is the man's acceleration relative to the block?
(b) As a result of the run, the block is forced to move right with a constant acceleration A (relative to the ground). What are ax and ay of the man, relative to the ground (in terms of A)?
(c) Find A.
(d) What's the block's displacement to the right during T?

Homework Equations





The Attempt at a Solution


(a)In the block's frame of reference and choosing a coordinate system which is parallel to the slope:
mgcos(α) = N; F - mgsin(α) = ma, where I chose F to denote the force exerted by the man running; Lcos(α) = 1/2*a*t2, which yielded a = 2Lcos(α)/T2
Are these equations correct? I am not sure.
(b) Won't the acceleration of the man now be:
a = [2Lcos(α)/T2 - Acos(α)]x + [N/m - gsin(α) - Asin(α)/m]y?
I am waiting for feedback on these first two sections before proceeding to the last two. Would truly appreciate some assistance in this.
 

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  • Diagram 3.jpg
    Diagram 3.jpg
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a) L is the running distance of the man, there is no additional cos(α) involved unless you want to consider the horizontal component of the acceleration only.

Forces and gravity are not relevant. Conservation of momentum is.
 

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