For Block A:

N=mg+Fsin30∘
=5×10+20×21
=60N
fr=(0.2)(60)=12N
Fcos30∘–fr=ma
2023 −12=5×a
⇒aA=5103−12 ……(i)
For Block B:

Fsin30∘+N=mg
N=5×10–20×(21)
=40N
fr=0.2×40=8N
Fcos30∘–fr=ma
20(23)−8=5×a
⇒aB=5103−8
aB−aA=54=0.8m/s2
A block of mass 5kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F=20N, making an angle of 30o with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is μ=0.2. The difference between the accelerations of the block, in case (B) and case (A) will be:
(g=10ms−2)

Held on 12 Apr 2019 · Verified 6 Jul 2026.
3.2ms−2
0ms−2
0.8ms−2
0.4ms−2
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