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Sav [38]
3 years ago
5

A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?

Physics
2 answers:
zysi [14]3 years ago
7 0
We know that -

P.E=m*g*h

Where,

m = mass

g = acceleration due to gravity

h=height

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

P.E.= 2*3000*9.8

P.E.= 58800 Joules

P.E.= 58.8 kJ

IRINA_888 [86]3 years ago
4 0
Sairah's work is correct as far as it goes.  The potential energy of the ball
relative to the bottom of the hill is 58,800 Joules.

<u>To address the second part of the question:</u>
In order to get ahold of that energy, the ball must be returned to the bottom
of the hill. The most efficient way would be to drop it, so that it wouldn't have to
scrape along the grass on the way down. But that can only work if there's a sheer
cliff on one side of the hill. Otherwise, you just have to roll it down, and accept the
fact that it loses some of its energy to friction on the way.

However the ball gets to the bottom, the energy it has left shows up in the form
of kinetic energy, and 58,800 joules is a lot of kinetic energy.  If somehow the ball
could arrive at the bottom with ALL the energy it had at the top, it would be moving
at something like 540 miles per hour !

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t=5

a=(v-v0)/t=(10-20)/5=-2

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A pole vaulter takes 1.3 seconds to fall from peak height to the landing mat. What is his vertical velocity at contact with the
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12.75 \frac{m}{s}

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6 0
2 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

5 0
2 years ago
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