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ss7ja [257]
3 years ago
15

A 4.0kg block is sliding with a constant velocity of 3.0m/s on a frictionless table that is 0.5m high. If all of the block’s ene

rgy were kinetic energy, how fast would the block be moving?
Physics
1 answer:
MArishka [77]3 years ago
8 0

Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

E_{M}=E_{p}  + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2}  \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]

All this energy will become kinetic energy and we can find the velocity.

37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]

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A woman lifts her 150 N child up one meter and carries her for a distance of 50 meters to the
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7500J

Explanation:

Given parameters:

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Unknown:

Work done by the woman = ?

Solution:

Work done by a body is the force applied to move a body through a given distance.

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 input the parameters:

   Work done = 150 x 50 = 7500J

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Given that the collision is elastic and glider 2 is initially at rest (v2,i =0), please use below Eqs. to explain why
Morgarella [4.7K]

Answer:

Explanation:

1 )

Put v2,i =0, in second equation

v2,f= (m2-m1)v2,i + 2m1v1,i/m1+m2

v2,f = 0 + 2m1v1,i/m1+m2

v2,f =  2m1v1,i/m1+m2

In this equation coefficient of v1,i is positive so v2,f and v1,i have the same sign.

2 )

Put m1 < m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

As m1-m2 is negative , v1f and v1i will have opposite sign.

3 )

Put m1 > m2  and v2,i =0 in first equation

v1,f= (m1-m2)v1,i + 2m2v2,1/m1+m2

v1,f= (m1-m2)v1,i

m1 - m2 is positive so v1f and v1i will have same  sign.

4 )

Put m1 = m2 and v2,i =0 in first equation

v1,f= (m1-m2)v1,i

= 0 because m1 = m2

So glider 1 will stop because v1,f = 0 .

 

 

5 0
2 years ago
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