Question #95639

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Question #95639

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Answer 1 · 2017-10-21T04:59:23

6m

Explanation:

When the Kinetic energy has reduced to half, the other part has become gravitational potential energy $E_{w} = m g h = \frac{58, 8 J}{2} = 29, 4 J$ $E_w=mgh=(58,8 J)/2= 29,4 J$
hence $h = \frac{E_{w}}{m g} = \frac{29, 4 J}{0, 5 k g \times 9, 8 \frac{m}{s^{2}}} = 6 m$ $h=E_w/(mg) = (29,4J)/(0,5kg xx 9,8m/s^2)=6m$

Answer 2 · 2017-10-21T05:01:24

$6 m$ $6"m"$

Explanation:

Initial Kinetic Energy $= {KE}_{i} = 58.8 J$ $="KE"_i=58.8J$

Final Kinetic Energy $= {KE}_{f} = \frac{58.8}{2} = 29.4 J$ $="KE"_f=58.8/2=29.4J$

Let's assume that the rest of Kinetic Energy has been converted into Gravitational potential energy.

Initial Gravitational potential Energy $= {PE}_{i} = 0 J$ $="PE"_i=0J$

Final Gravitational potential Energy $= {PE}_{f} = 58.8 - 29.4 = 29.4 J$ $="PE"_f=58.8-29.4=29.4J$

We need to know the height $h$ $color(blue)h$

So, ${PE}_{f} = m g h$ $"PE"_f=mgcolor(blue)h$

$\Rightarrow 29.4 J = 0.5 kg \times 9.8 \frac{m}{s^{2}} \times h$ $=>29.4J=0.5"kg"xx9.8m/s^2xxcolor(blue)h$

$\Rightarrow h = \frac{29.4 J}{0.5 kg \times 9.8 \frac{m}{s^{2}}}$ $=>color(blue)h=(29.4J)/(0.5"kg"xx9.8m/s^2)$

$=>color(blue)h=(29.4cancelN m)/(0.5cancel"kg"xx9.8cancel(m/s^2))$

$color(blue)h=6m$

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Question #95639

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