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Rina8888 [55]
2 years ago
5

Patrick is repairing the roof on his house. He uses a hammer having a mass of 1 kilogram. While working at the apex of the roof,

the hammer slips and falls to the ground at a velocity of 12 meters/second. Find the potential energy of the hammer just before slipping.
Physics
1 answer:
pav-90 [236]2 years ago
5 0

Answer:

72 joules

Explanation:

The potential energy of that hammer is a function of its displacement against gravity. Considering that it fell with a velocity of 12 m/s, it was its displacement against gravity that gave it this velocity. It will continue to move until its displacement to gravity is zero.

since the body is in motion; it has converted its potential energy (mgh, m is mass, g is acceleration due to gravity, and h is the height) to kinetic energy (energy due to motion, 1/2mv^2; m = mass, v = velocity or speed)

therefore the potential energy is equal to kinetic energy

mgh = 1/2mv^2 = 1/2 *1kg* 12*12 = 72 joules.

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Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre
castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:  

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)  

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619  

=

t

+

C

Here C =constant of integration.   

x

=

0  at  t

=

0

, we get:  C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619  =

t

−

4.7619

.

.

.

.

.

.

.

.

.

(

2

)

Here

x  is measured in miles and  t  in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),    

     to get:

      = −

4.7619  

      =  

1

−

4.7619

      = −

3.7619

  or  x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in   m

i

l

/

differentiate  

     equation (1) with respect to time.

     we have to eliminate x from the equation (1) using equation (2).  

     Eliminating x we get:

     v

=

7.5×

     Now differentiating above equation w.r.t time we get:

      a

=

d

v/

d

t

       =

−

0.675

/

      At  

      t

=

0

      the joggers acceleration is :

       a

=

−

0.675

m

i

l

/

        =

−

4.34

×

f

t

/  

(c)  required time for the jogger to run 6 miles is obtained by setting  

        x

=

6  in equation (2).  We get:

        −

4.7619

(

1

−

(

0.04

×

6  )

)^

7

/

10=

t

−

4.7619

         or

         t

=

0.832

h

r

s

6 0
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