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Sagot :
Answer:
Look at explanation
Explanation:
a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2
b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2. A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.
The relation will be:
(a) K1 = K2
(b) K1 < K2
According to the question,
- Potential energy be "P".
- Kinetic energy be "K".
(a)
Word done towards both the block will be similar.
So,
→ [tex]P1 = P2[/tex]
→ [tex]K1= K2[/tex]
(b)
We know,
→ [tex]a = \frac{F}{M}[/tex]
or,
→ [tex]V = a\times t[/tex]
Now,
→ [tex]K = \frac{1}{2} MV^2[/tex]
[tex]= 0.5\times M\times V^2[/tex]
[tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]
[tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]
The force and t will be same. So K of the smaller mass will be greater than the larger mass.
hence,
→ [tex]K1<K2[/tex]
Thus the above responses are correct.
Learn more about friction here:
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