A block slides on a table pulled by a string attached to a hanging weight. In case 1 the block slides without friction and in ca
2 answers:
Response:
In the second scenario
Clarification:
Reasoning:
Friction opposes motion.
In scenario one, the block glides without any friction present, indicating a smooth surface allowing minimal net force to pull the string.
In the second scenario, kinetic friction plays a role between the sliding block and the table, where kinetic friction is expressed as:
with relevant variables being as the coefficient of friction and N as the normal force.
This means the necessary net force becomes:
Thus, the force is greater in the second situation.
Consequently, the tension is also heightened in the second scenario.
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Answer:
A) denotes the resultant velocity of cart B post-collision.
B)
C)
D)
E)
F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.
G) Yes, momentum is conserved in every elastic collision.
Explanation:
Given:
- mass of car A,
- mass of car B,
- initial velocity of car A,
- final velocity of car A,
A)
The question mentions the cars experience an elastic collision:
By applying momentum conservation principles:
denotes the resulting velocity of cart B after collision.
B)
Initial kinetic energy of cart A:
C)
Initial kinetic energy of cart A:
D)
The final kinetic energy of cart A:
E)
The final kinetic energy of cart B:
F)
Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.
G)
Indeed, momentum is consistently conserved in elastic collisions.
Answer:
0.0984
Explanation:
The first diagram below illustrates a free body diagram that will aid in resolving this problem.
According to the diagram, the force's horizontal component can be expressed as:
Substituting 42° for θ and 87.0° for
Meanwhile, the vertical component is:
Again substituting 42° for θ and 87.0° for
In resolving the vector, let A denote the components in mutually perpendicular directions.
The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ
The frictional force can be expressed as:
Where;
is the coefficient of friction
N = the normal force
Also, the normal reaction (N) is calculated as mg - F sin θ
Substituting . Normal reaction becomes:
By balancing the forces, the horizontal component of the force equals the frictional force.
The horizontal component is described as follows:
Rearranging the equation above to isolate leads to:
Substituting in the following values:
m = 73 kg
g = 9.8 m/s²
Thus:
Therefore, the coefficient of friction is = 0.0984
