Answer:
a.25J b.6J c.6J d.31J e.[tex]\sqrt{31}[/tex] m/s
Explanation:
a. using the formula Ek=1/2mv^2, the kinetic energy would be 1/2*2*5^2=25J
b. we can separate the work the child applied. So the work he/she applied would be 2*3=6J (W=FS)
c. by the "Energy conservation law", we knew the energy can be "add up" since they wouldn't "disappear". Thus, the change in the car’s kinetic energy is 6J.
d.25+6=31J
e.now you know the new Ek, and the m, just use the formula again! 31=1/2*2*v^2. So the velocity after the child push it would be [tex]\sqrt{31}[/tex] m/s
The kinetic energy and the work done by the car is:
(a) 25 J
(b) 6 J
(c) 6 J
(d) 31 J
(e) 5.57 m/s
According to the question,
Force,
Distance,
(a)
The Kinetic energy of the car will be:
→ [tex]K_i = \frac{1}{2} mv_0^2[/tex]
[tex]= \frac{1}{2}\times (2)\times (5)^2[/tex]
[tex]= 25 \ J[/tex]
(b)
The work done by the car will be:
→ [tex]W = fd[/tex]
[tex]= 3\times 2[/tex]
[tex]= 6 \ J[/tex]
(c)
The change in the car's K.E will be:
→ [tex]\Delta K = W[/tex]
[tex]= 6 \ J[/tex]
(d)
The car's final K.E will be:
→ [tex]K_f = W+K_i[/tex]
[tex]= 6+25[/tex]
[tex]= 31 \ J[/tex]
(e)
The final velocity of the car will be:
→ [tex]v_f = \sqrt{\frac{2 K_f}{m}}[/tex]
[tex]= \sqrt{\frac{2\times 31}{2} }[/tex]
[tex]= 5.57 \ m/s[/tex]
Thus the above solution is correct.
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