Work

Question 1

The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the breaks on cars A and B respectively, then the ratio FA/FB is

Accepted Answer

2/3. By the work-energy theorem the braking force does work equal to the entire initial kinetic energy, so F = K/s. Hence FA/FB = (KA/sA) x (sB/KB) = (100/1000) x (1500/225) = 2/3.

Answer

3/2

Answer

1/3

Answer

1/2

Question 2

A particle of mass M moves along a horizontal x axis from x = 0 to x = L. The coefficient of kinetic friction varies as a function of x as μk(x) = μ₀ − αx, where μ₀ and α are constants of appropriate dimensions, so that μk(L) = 0. The total work done by the frictional force during the motion is n μ₀MgL, where g is the acceleration due to gravity. The value of n is:

Accepted Answer

1/2. Friction is not constant here, so the work is an integral rather than a product: |W| = ∫₀ᴸ μk(x) Mg dx = Mg ∫₀ᴸ (μ₀ − αx) dx. The condition μk(L) = 0 fixes α = μ₀/L, and the integral then gives μ₀MgL − μ₀MgL/2 = μ₀MgL/2, so n = 1/2.

Answer

3

Answer

1

Answer

1/3

Question 3

At any instant of time t, the displacement of a particle is given by 2t − 1 (SI unit) under the influence of a force of 5 N. The value of instantaneous power is (in SI unit)

Accepted Answer

10. Instantaneous power is the product of force and instantaneous velocity. Differentiating the displacement x = 2t − 1 gives v = dx/dt = 2 m s⁻¹, a constant. So P = Fv = 5 × 2 = 10 W. The −1 in the displacement is a constant offset and disappears on differentiation, which is what the distractors are built around.

Answer

5

Answer

7

Answer

6

Q1.The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If FA and FB are the forces applied by the breaks on cars A and B respectively, then the ratio FA/FB is

Q3.At any instant of time t, the displacement of a particle is given by 2t − 1 (SI unit) under the influence of a force of 5 N. The value of instantaneous power is (in SI unit)