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3 years ago

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A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that

is 12 m from the point of impact when the key is re- leased. What is the speed of the boat

1 answer:

guapka [62]3 years ago

7 0

Answer:

4m/s

Explanation:

The motion is in downward direction, and the boat is moving with constant velocity, the position of the boat horizontally is 12m before the key fall.

The height difference "h"= ( y- y°)= 45 m

We can determine the needed time the key requires to reach the water using the expression from Newton's law

h= v(o) + 1/2 at^2

g= acceleration due to gravity= 9.8m/s^2

h= v(o) + 1/2 gt^2 -------------------------eqn*)

V(o)= 0 for the key= initial velocity

h= height that the key falls= 45m

If we substitute the above values into eqn(*) we have

We can make t^2 subject of the formula since v(0)= 0

t^2= h/(1/2 g)

= 45/(1/2× 9.8)

t^2= 45/4.9=9.18

t= √9.18

t=3.03 secs

Since the boat was moving with constant velocity,

Then Velocity= distance/ time

= 12/3.03

Velocity= 4m/s

Hence, the speed of the boat is 4m/s

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Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

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d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

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$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

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The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

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$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

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$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

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$\Delta K = \Delta W$ (5)

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$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

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d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

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3 years ago

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