Student sliding bag along floor in elevator

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  • #1
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Homework Statement
Pls see below
Relevant Equations
Pls see below
For this problem,
1676877869180.png

The solution is,
1676877943337.png

However, is the reason why they don't do ##-0.5a_xt^2## because the negative of the acceleration has already taken care of itself?

Many thanks!
 
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  • #2
The answer seems to make
[tex]a_x=-\alpha_k (g+a)<0[/tex]
negative. You can make it positve one with your setting of ##-0.5a_xt^2##.
 
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  • #3
anuttarasammyak said:
The answer seems to make
[tex]a_x=-\alpha_k (g+a)<0[/tex]
negative. You can make it positve one with your setting of ##-0.5a_xt^2##.
Than you for your reply @anuttarasammyak !
 
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FAQ: Student sliding bag along floor in elevator

What is the effect of the elevator's motion on the friction between the bag and the floor?

The motion of the elevator can affect the normal force, which in turn affects the friction between the bag and the floor. If the elevator is accelerating upwards, the normal force increases, leading to greater friction. Conversely, if the elevator is accelerating downwards, the normal force decreases, resulting in less friction.

How does the direction of the elevator's acceleration influence the sliding motion of the bag?

If the elevator is accelerating upwards, the increased normal force can make it harder for the student to slide the bag due to increased friction. If the elevator is accelerating downwards, the reduced normal force can make it easier to slide the bag as friction decreases.

Does the speed of the elevator affect the sliding bag's motion?

The constant speed of the elevator (whether moving up or down) does not affect the sliding motion of the bag. Only changes in the elevator's acceleration will impact the normal force and thus the friction between the bag and the floor.

What happens to the bag if the elevator suddenly stops?

If the elevator suddenly stops, the inertia of the bag will cause it to continue moving at its current velocity. If the elevator was moving upwards and stops suddenly, the bag may briefly experience a decrease in normal force, potentially reducing friction momentarily.

How can we calculate the frictional force acting on the bag in a moving elevator?

The frictional force can be calculated using the formula \( F_f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. In a moving elevator, the normal force \( N \) is affected by the elevator's acceleration and can be calculated as \( N = m(g + a) \) for upward acceleration or \( N = m(g - a) \) for downward acceleration, where \( m \) is the mass of the bag, \( g \) is the acceleration due to gravity, and \( a \) is the elevator's acceleration.

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