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Laws of Motion question

2019 · 12 Apr · Shift 2 · Q48
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Laws of Motion question

2019 · 12 Apr · Shift 2 · Q48

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is μ\muμ = 0.2. The difference between the accelerations of the blocks, in case (B) and case (A) will be : (g = 10 ms–2) JEE Main 2019 (Online) 12th April Evening Slot Physics - Laws of Motion Question 104 English
  1. A
    3.2 ms–2
  2. B
    0.8 ms–2
  3. C
    0 ms–2
  4. D
    0.4 ms–2
View written solutionFree

Correct answer: B

  1. Given data
  • Mass of block: m=5 kgm = 5\,\text{kg}m=5kg
  • Applied force: F=20 NF = 20\,\text{N}F=20N
  • Angle with horizontal: θ=30∘\theta = 30^\circθ=30∘
  • Coefficient of friction: μ=0.2\mu = 0.2μ=0.2
  • Acceleration due to gravity: g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

Also,

Fcos⁡30∘=20⋅32=103 NF\cos 30^\circ = 20\cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}\,\text{N}Fcos30∘=20⋅23​​=103​N Fsin⁡30∘=20⋅12=10 NF\sin 30^\circ = 20\cdot \frac{1}{2} = 10\,\text{N}Fsin30∘=20⋅21​=10N

The horizontal component produces motion, while the vertical component changes the normal reaction and hence friction.

  1. Case (A): Block is pushed

When the block is pushed at 30∘30^\circ30∘ downward to the horizontal, the vertical component acts downward.

So normal reaction is:

NA=mg+Fsin⁡30∘=5⋅10+10=60 NN_A = mg + F\sin 30^\circ = 5\cdot 10 + 10 = 60\,\text{N}NA​=mg+Fsin30∘=5⋅10+10=60N

Friction is:

fA=μNA=0.2×60=12 Nf_A = \mu N_A = 0.2\times 60 = 12\,\text{N}fA​=μNA​=0.2×60=12N

Horizontal net force:

Fnet,A=Fcos⁡30∘−fA=103−12F_{\text{net},A} = F\cos 30^\circ - f_A = 10\sqrt{3} - 12Fnet,A​=Fcos30∘−fA​=103​−12

Hence acceleration:

aA=103−125a_A = \frac{10\sqrt{3} - 12}{5}aA​=5103​−12​

Using 3≈1.732\sqrt{3} \approx 1.7323​≈1.732,

103≈17.3210\sqrt{3} \approx 17.32103​≈17.32 aA=17.32−125=5.325=1.064 m s−2a_A = \frac{17.32 - 12}{5} = \frac{5.32}{5} = 1.064\,\text{m s}^{-2}aA​=517.32−12​=55.32​=1.064m s−2
  1. Case (B): Block is pulled

When the block is pulled at 30∘30^\circ30∘ upward to the horizontal, the vertical component acts upward.

So normal reaction is:

NB=mg−Fsin⁡30∘=50−10=40 NN_B = mg - F\sin 30^\circ = 50 - 10 = 40\,\text{N}NB​=mg−Fsin30∘=50−10=40N

Friction is:

fB=μNB=0.2×40=8 Nf_B = \mu N_B = 0.2\times 40 = 8\,\text{N}fB​=μNB​=0.2×40=8N

Horizontal net force:

Fnet,B=Fcos⁡30∘−fB=103−8F_{\text{net},B} = F\cos 30^\circ - f_B = 10\sqrt{3} - 8Fnet,B​=Fcos30∘−fB​=103​−8

Hence acceleration:

aB=103−85a_B = \frac{10\sqrt{3} - 8}{5}aB​=5103​−8​

Numerically,

aB=17.32−85=9.325=1.864 m s−2a_B = \frac{17.32 - 8}{5} = \frac{9.32}{5} = 1.864\,\text{m s}^{-2}aB​=517.32−8​=59.32​=1.864m s−2
  1. Difference between accelerations
aB−aA=1.864−1.064=0.8 m s−2a_B - a_A = 1.864 - 1.064 = 0.8\,\text{m s}^{-2}aB​−aA​=1.864−1.064=0.8m s−2

So the required difference is:

0.8 m s−2\boxed{0.8\,\text{m s}^{-2}}0.8m s−2​
  1. Option check
  • A: 3.2 m s−23.2\,\text{m s}^{-2}3.2m s−2 — incorrect
  • B: 0.8 m s−20.8\,\text{m s}^{-2}0.8m s−2 — correct
  • C: 0 m s−20\,\text{m s}^{-2}0m s−2 — incorrect
  • D: 0.4 m s−20.4\,\text{m s}^{-2}0.4m s−2 — incorrect

Therefore, the correct option is B.

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