Question

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.

Answer 1

Answer:

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

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

$KE_A >KE_B$

Therefore , option A is correct .