A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8
N. A normal force (n->) and a friction force (f->) also act on the box. What is the magnitude of the friction force?
74 N
2.4 N
69 N
4.8 N
9.6 N
74 N
2.4 N
69 N
4.8 N
9.6 N
1 answer:
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Answer:
4. both blocks will both have the same amount of kinetic energy.
Explanation:
When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.
Spring force is given as:
where: length of compression in the spring
<u>We know kinetic energy is given by:</u>
Hence the kinetic energy of both the blocks is equal when they are released to move free.
To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.
By definition kinetic energy is defined as
Where,
m = Mass
v = Velocity
On the other hand we have the conservation of the moment, which for this case would be defined as
Here,
m = Total mass (8Kg at this case)
Mass each part
Initial velocity
Final velocity particle 2
Final velocity particle 1
The initial kinetic energy would be given by,
In the end the energy increased 100J, that is,
By conservation of the moment then,
Replacing we have,
(1)
In the final point the cinematic energy of EACH particle would be given by
(2)
So we have a system of 2x2 equations
Replacing (1) in (2) and solving we have to,
200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)
PART A:
Then replacing in (1) we have that
PART B: