The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
<h3>
Acceleration of the cart</h3>
The acceleration of the cart is determined from the net force acting on the mass-cart system.
Upward force = Downward force
ma = mg
13a = 10(9.8)
13a = 98
a = 98/13
a = 7.54 m/s²
Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
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Explanation:
(a) Net force acting on the block is as follows.

or, ma = -mg Sin (\theta)[/tex]
a = 
= 
= -3.35 
According to the kinematic equation of motion,

Distance traveled by the block before stopping is as follows.
s = 
= 
= 21.5 m
According to the kinematic equation of motion,
v =
0 = 
= 7.16 sec
Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.
(b) When the block is moving down the inline then net force acting on the block is as follows.

ma = 
a = 
= 
= 3.35 
Kinematics equation of the motion is as follows.
s = 
21.5 m = 
= 
= 3.58 sec
Hence, total time taken by the block to return to its starting position is as follows.
t = 
= 7.16 sec + 3.58 sec
= 10.7 sec
Thus, we can conclude that 10.7 sec time it take to return to its starting position.
You test it over and over again. so basically you experiment on it more.
Hope this helps!
Answer:
a ut will move faster than the large object was moving initially