Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2.
(a) When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (Select all that apply.)
K1 < K2 p1 = p2 p1 < p2 p1 > p2 K1 > K2 K1 = K2
(b) When a force is applied to object 1, it accelerates for a time interval ?t. The force is removed from object 1 and is applied to object 2. Which statements are true after object 2 has accelerated for the same time interval ?t? (Select all that apply.)
K1 > K2 K1 = K2 p1 = p2 p1 > p2 K1 < K2 p1 < p2

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.

Cricetus

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:

https://brainly.com/question/13340887

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.