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tensa zangetsu [6.8K]
3 years ago
14

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

conservation of energy to show that the speed of the banana just before hitting the water is v
Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

              Total Energy before fall = Total Energy after fall

                                E_{i}  = E_{f}

Here, total energy is the sum of kinetic energy and potential energy

K.E_{i} + P.E_{i} = K.E_{f} + P.E_{f}       (a)

When banana is at height h, it has

                 K.E_{i} = 0    and    P.E_{i} = mgh          

and when it reaches the river, it has

       K.E_{f}  = 1/2mv^{2}    and   P.E_{f}  = 0

Putting the values in equation (a)

                              0 + mgh = 1/2mv^{2} + 0

                                      mgh = 1/2mv^{2}

<em>cutting 'm' from both sides</em>

<em>                                           </em>gh = 1/2v^{2}

                                          v = \sqrt{2gh}

Hence, the velocity of banana before hitting the water is

                                          v = \sqrt{2gh}

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A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

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Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

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F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

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e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

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3 years ago
What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a
Naya [18.7K]

Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

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Now horizontal component of tension

Tx= T cos 40 = T× 0.7660  N

==>T× 0.7660 = 29.4

==> T= 38.38 N

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