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Nadya [2.5K]
3 years ago
7

A ball of mass 0.35 kg hangs straight down on a string of length 13 cm. It is then swung upward, keeping the string taut, until

the string makes an angle of 70° with respect to the vertical. Find the change in the gravitational potential energy of the Earth-ball system.

Physics
1 answer:
Anni [7]3 years ago
6 0

Answer:

PE=0.29J

Explanation:

According to the description, there is a angle and in point swung upward of 70°

So,

Y=13*10^{-2}*cos(70) \\Y=0.0444m

Appling the equation of Potential Energy we have,

PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J

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Answer:

<h3>The coefficient of kinetic friction between the puck and the ice is \mu _{k} = 0.12</h3>

Explanation:

Given :

Initial speed  v_{o} = 9.5 \frac{m}{s}

Displacement x = 37.4 m

From the kinematics equation,

  v^{2} - v^{2} _{o}  = 2ax

Where v^{2}   = final velocity, in our example it is zero (v =0), a = acceleration.

   a =- \frac{90.25}{ 2 \times 37.4}

   a =- 1.21 \frac{m}{s^{2} }

From the formula of friction,

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Minus sign represent friction is oppose the motion

Where N = mg ( normal reaction force )

 ma = -\mu _{k}  m g                                                  ( ∵ g = 9.8 \frac{m}{s^{2} } )

So coefficient of friction,

 \mu_{k} = \frac{1.21}{9.8}

 \mu_{k} = 0.12

Therefore, the coefficient of kinetic friction between the puck and the ice is  \mu _{k} = 0.12 .

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