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Aleksandr-060686 [28]
3 years ago
13

An elevator is being pulled up from the ground floor to the third floor by a cable. The cable is exerting 4500 newtons of force

on the elevator. According to Newton's law of action-reaction, what is the gravitational force on the elevator, in newtons, if the elevator is NOT accelerating?
Physics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

The gravitational force on the elevator = 4500N

Explanation:

The given parameters are;

The force applied by the elevator, F  = 4500 N

The acceleration of the elevator = Not accelerating

From Newton's third law of motion, the action of the cable force is equal to the reaction of the gravitational force on the elevator which is the weight, W and motion of the elevator as follows;

F = W + Mass of elevator × Acceleration of elevator

∴ F = W + Mass of elevator × 0 = W

F = 4500 N = W

The net force on the elevator is F - W = 0

The gravitational force on the elevator = W = 4500N.

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
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Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

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-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

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