JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Laws of Motion

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

3 0^{\circ}

with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is

m = 0.2

. The difference between the accelerations of the block, in case (B) and case (A) will be : (

g = 10 ms^{- 2}

)

Answer: (D)

Solution

Pushing the force down at

3 0^{\circ}

presses the block harder onto the floor:

N_{A} = m g + F sin 3 0^{\circ} = 50 + 10 = 60

N, so friction is

0.2 \times 60 = 12

N and the net drive is

F cos 3 0^{\circ} - 12 = 17.3 - 12 = 5.3

N, giving

a_{A} = \frac{5.3}{5} \approx 1.06

m/s

^{2}

. Pulling up at

3 0^{\circ}

lightens the floor:

N_{B} = m g - F sin 3 0^{\circ} = 50 - 10 = 40

N, friction is

8

N, the net drive is

17.3 - 8 = 9.3

N and

a_{B} = \frac{9.3}{5} \approx 1.86

m/s

^{2}

. The difference is

a_{B} - a_{A} \approx 0.8

m/s

^{2}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option