JEE Challenger
More from Laws of Motion
PhyMedAdv 2007 • Paper 1

Mechanical Energy Loss for Block Moving on Inclined Plane

STATEMENT-1
A block of mass mm starts moving on a rough horizontal surface with a velocity vv. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30∘30^\circ with the horizontal and the same block is made to go up on the surface with the same initial velocity vv. The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
because
STATEMENT-2
The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

Options

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

Correct
D

Statement-1 is False, Statement-2 is True

Topics & Concepts

Laws of MotionFrictionWork-Energy Theorem

Step-by-Step Solution

To evaluate the statements:

  1. Analysis of Statement-1:

    • First Situation (Horizontal surface):
      The block starts with an initial kinetic energy Ki=12mv2K_i = \frac{1}{2}mv^2 and potential energy Ui=0U_i = 0.
      It eventually comes to rest on the horizontal surface, so its final kinetic energy is Kf=0K_f = 0 and its potential energy remains Uf=0U_f = 0.
      The decrease in mechanical energy is:
      ΔE1=Ei−Ef=12mv2−0=12mv2\Delta E_1 = E_i - E_f = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2
      All of this mechanical energy is dissipated as heat by the work done against friction.

    • Second Situation (Inclined surface at θ=30∘\theta = 30^\circ):
      The block is projected up the incline with the same initial kinetic energy Ki=12mv2K_i = \frac{1}{2}mv^2 (taking initial potential energy Ui=0U_i = 0).
      The retardation of the block moving up the incline is:
      a=gsin⁡θ+μgcos⁡θa = g\sin\theta + \mu g\cos\theta
      Using the kinematic relation v2=2asv^2 = 2as, the distance ss moved by the block along the incline before coming to rest is:
      s=v22(gsin⁡θ+μgcos⁡θ)s = \frac{v^2}{2(g\sin\theta + \mu g\cos\theta)}
      The decrease in mechanical energy of the system is equal to the work done against the non-conservative friction force:
      ΔE2=Wfriction=fk⋅s=(μmgcos⁡θ)⋅v22(gsin⁡θ+μgcos⁡θ)=12mv2(μcos⁡θsin⁡θ+μcos⁡θ)\Delta E_2 = W_{\text{friction}} = f_k \cdot s = (\mu mg\cos\theta) \cdot \frac{v^2}{2(g\sin\theta + \mu g\cos\theta)} = \frac{1}{2}mv^2 \left(\frac{\mu\cos\theta}{\sin\theta + \mu\cos\theta}\right)
      Since sin⁡30∘=12>0\sin 30^\circ = \frac{1}{2} > 0, we have:
      μcos⁡θsin⁡θ+μcos⁡θ<1  ⟹  ΔE2<12mv2=ΔE1\frac{\mu\cos\theta}{\sin\theta + \mu\cos\theta} < 1 \implies \Delta E_2 < \frac{1}{2}mv^2 = \Delta E_1
      Part of the initial mechanical energy is converted into gravitational potential energy instead of being dissipated. Therefore, the decrease in mechanical energy in the second situation is indeed smaller than in the first situation.

    Thus, Statement-1 is True.

  2. Analysis of Statement-2:

    • The coefficient of friction (μ\mu) depends only on the material properties and roughness of the surfaces in contact. It is completely independent of the angle of inclination of the plane.

    Thus, Statement-2 is False.

Therefore, Statement-1 is True and Statement-2 is False.

Correct Option: C

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