A block oscillating up and down on a spring comes to the top of its path, where it is momentarily at rest. At the instant it is
1 answer:
Opposite to the direction of the velocity which led it to its current position.
Explanation:
The direction of momentum when a vertically oscillating block comes to the rest momentarily will be opposite to the direction of the velocity that it has just followed to reach reach its current position.
The direction of change in momentum at the bottom will be upwards and at the top will be downwards.
The change in momentum is mathematically defined as:
where:
mass of the block
final velocity of the block
initial velocity of the block
When the block comes to rest it is due to the result of continuously decreasing velocity.
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Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get
10N = 50 kg · a
A newton is just
The s^2 and 50 kg you multiply
The kg's cancel and 10/50 is 1/5
So the acceleration is 1/5 m/s^2
Answer: Option B: 1.3×10⁵ W
Explanation:
Work Done,
Where s is displacement in the direction of force and F is force.
where, v is the velocity.
It is given that, F = 5.75 × 10³N
v = 22 m/s
P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W
Thus, the correct option is B