A constant force of 20 N is acting on a body of mass 2 kg which is at rest on a rough
initially. Find the acceleration produced and the distance covered in 5 s. Coefficient of
is 0.5.
a) 4.1 ms-2, 62.75 m b) 5.1 ms-2, 67.33 m ) 5.1 ms-2, 63.75 m d) 4.5 ms-2, 601
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Answered by
0
Answer:
10m/s^2=acc. Distance =50 m
Answered by
2
Answer:
acceleration = 5 m/s²
distance covered = 62.5 m
Explanation:
net force = applied force - friction
friction = coefficient × normal
= 0.5 × (2 × 10)
= 10 N
•°•, net force = 20 - 10 = 10N
acceleration = net force ÷ mass of body
= 10 ÷ 2
= 5 m/s²
distance covered = ½ × a × t²
= ½ × 5 × 5²
= 125 ÷ 2
= 62.5 m
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