Numerical Problem based on Newton's laws of Motion

In summary, the 5kg block on top of the 10kg block will slide off completely in 2 seconds if the 10kg block is accelerating at 5m/s^2 and there is no friction between the blocks. This is found by using the equations of motion, with the length of the top of the 10kg block being 10m.
  • #1
konichiwa2x
81
0
A 5kg block is resting on the top right hand corner of a 10kg block. The length of the top of the 10kg block is 10m. Find the time taken by the 5kg block on top to slide of the 10kg block completely if the 10kg block is accelerating at 5m/s^2.

My work:

aB = accleration of the 5kg block
aA = acceleratipon of the 10kg block
If the 10kg block is accelerating to the right, then a pseudo force must be acting on the 5kg block on it in the opposite direction,

so fp = 5aB
aB = 5

aB - aA = -10

-10 = 1/2(-10)t^2
t = (2)^1/2

where am i going wrong? My book says 2 seconds is the correct answer. Please help:confused:
 
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  • #2
Since there is no friction in the problem, there is no force acting on the small block. Your "pseudo" acceleration should simply be the opposite direction of the acceleration of the large block. Your error is in adding the 2 accelerations. If you just use 5 [itex] \frac m {s^2} [/tex] as the acceleration the result is 2s.
 
  • #3
ok so this pseudo acceleration is just the acceleration of the block B with respect to the block A, whereas the actual acceleration of the block B is 0. is that correct?
 
  • #4
since the lower block is of 10 kg and the upper block is just half its weight so the dimensions of the upper block will be half the lower block. so the length of the upper block is 5m.
now since there is no friction the lower block will have to move 10m so that tha upper block is completely fallen away.
now S = ut+1/2at^2
here u=0
a=5
s=10
so, 10=1/2 5t^2
t^2=4
t=2sec.
 
  • #5


Your calculations are correct, but you have not taken into account the initial velocity of the 5kg block. Since it is initially at rest, its initial velocity is 0 m/s. Using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can find the final velocity of the 5kg block when it slides off the 10kg block.

v = u + at
v = 0 + (5)(2)^1/2
v = 5(2)^1/2 m/s

Now, using the formula s = ut + 1/2at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time, we can find the distance traveled by the 5kg block before it slides off the 10kg block.

s = ut + 1/2at^2
s = (0)(2)^1/2 + 1/2(5)(2)^1/2(2)^1/2
s = 5 m

Since the length of the top of the 10kg block is 10m, the 5kg block will slide off completely after 2 seconds (10m/5m/s = 2 seconds). Therefore, the correct answer is indeed 2 seconds.
 

1. What are Newton's laws of motion?

Newton's laws of motion are three fundamental principles proposed by Sir Isaac Newton in his book "Philosophiæ Naturalis Principia Mathematica" in 1687. They describe the relationship between the forces acting on an object and the motion of that object.

2. How can Newton's laws of motion be applied to numerical problems?

Newton's laws of motion can be used to solve numerical problems by understanding the forces acting on an object and using the equations derived from the laws to calculate the resulting motion. This can be done by breaking down the problem into smaller parts and applying the laws step by step.

3. What is the first law of motion?

The first law of motion, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will continue in motion at a constant velocity unless acted upon by an external force.

4. How is the second law of motion used in numerical problems?

The second law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. This law can be used in numerical problems by using the equation F=ma to calculate the net force on an object and its resulting acceleration.

5. What is the third law of motion?

The third law of motion states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object will exert an equal and opposite force back. This law can be used in numerical problems by considering the forces acting on multiple objects and their resulting interactions.

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