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chubhunter [2.5K]
2 years ago
8

What is the gain in gravitational potential energy of a body of weight 2000N as it rises from a height of 20m to a height of 25

m above the Earth's surface
Physics
1 answer:
monitta2 years ago
8 0

Answer:

Explanation:

GPE =weight * height

       = 2000*(25-20)

      =  10000 J

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How can I answer question 19? please help me, I need the answer now
Crank
The spring starts out 22 cm long with nothing hanging on it.

Hanging 35 newtons of weight on it stretches the spring 1 meter.

Ellen is going to hang 250 grams of mass on the spring.
What's the weight of 250 grams of mass ?

   Weight = (mass) x (acceleration of gravity in the place where the mass is) .

On Earth, the acceleration of gravity is 9.8 m/s² .
250 grams is 0.25 of a kilogram.

Weight of 250 grams  =  (0.25 kilogram) x (9.8 m/s²)

                                 =    (0.25 x 9.8)  kg-m/s²

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6 0
2 years ago
What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?
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Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

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Let the masses of two bodies are denoted as m_{1} \ and\ m_{2}\ respectively

Let\ m_{1} =1 kg\ and\ m_{2} =2 kg

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

                             m_{1} g-T=m_{1} a        [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction [\mu]=0.13

Hence\ the\ frictional\ force\ F=\mu N

                                                     F=\mu m_{2} g

Hence the net force acting on the body having mass 2 kg-

                                  T-\mu m_{2} g=m_{2} a  [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

                           m_{1} g-T=m_{1}a             [1]

                           T-\mu m_{2}g= m_{2} a   [2]

                           ---------------------------------------------------

                           [m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a

                           a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g

                           a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2

                           a=\frac{0.74}{3} *9.8\ m/s^2

                           a=2.417 m/s         [ans]

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