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Tasya [4]
3 years ago
14

Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel

to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______
a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.
Physics
1 answer:
krek1111 [17]3 years ago
6 0

Answer:

The correct answer is <u>option (A) that is KEA > KEB .</u>

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

W=F_d- F_f_r_id-F_gh

W=F_d-\mu_kmgdcos\theta-mgdsin\theta

The change in kinetic energy is ,

   \Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2

At the top of the inclined plane , the velocity is zero

So,

\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2

\Delta KE=-\frac{1}{2}m\nu_0^2

From the work energy theorem , we have W=-\Delta K in case of friction , so

\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta

KE=Fd-\mu_kmgdcos\theta-mgdsin\theta

For object A-

KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta

For object B

KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta

KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

KE_A >KE_B

<u>Therefore , option A is correct .</u>

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