How much force would be needed to cause a 4.6kg object to accelerate at 9.2m/s/s? *
1 answer:
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Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
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Answer:
The coefficient of kinetic friction between the sled and the snow is 0.0134
Explanation:
Given that:
M = mass of person = 52 kg
m = mass of sled = 15.2 kg
U = initial velocity of person = 3.63 m/s
u = initial velocity of sled = 0 m/s
After collision, the person and the sled would move with the same velocity V.
a) According to law of momentum conservation:
Total momentum before collision = Total momentum after collision
MU + mu = (M + m)V
Substituting values:
The velocity of the sled and person as they move away is 2.81 m/s
b) acceleration due to gravity (g) = 9.8 m/s²
d = 30 m
Using the formula:
The coefficient of kinetic friction between the sled and the snow is 0.0134