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Answer:
<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>
Explanation:
<u>Net Force
</u>
The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.
The acceleration is proportional to the net force and inversely proportional to the mass of the object.
If the object has zero net force, it won't get accelerated and its velocity will remain constant.
The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.
Since there is an external force applied to the box, it must have been balanced by the force of friction, thus the force of friction has the same magnitude acting opposite to the applied force.
The force of friction acting on the block has a magnitude of 15 N opposite to the applied force.
Answer:
46.67 N/m
Explanation:
mass, m = 0.1 kg
distance, y = 2.1 cm = 0.021 m
Let K be the spring constant.
F = mg = Ky
0.1 x 9.8 = K x 0.021
K = 46.67 N/m
Thus, the spring constant of the spring is 46.67 N/m.
The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
where:
is the momentum of block A before the collision
is the momentum of block B before the collision
is the momentum of block A after the collision
is the momentum of block B after the collision
Solving for , we find:
So, the momentum of block B after the collision is -50 kg m/s.
Learn more about momentum:
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