Question

0)
A body of mass 0.5 kg starts from rest (t=0) & experiences a force F= 81 +6]. What will be its
velocity at t=1 if it covers a distance of 4m?​

Answer 1

Explanation:

To solve this problem we need to assume a units for mass of the body and a unit for force. Let these be kilogram (= kg) and Newton (= N).

Mass of body = 0.5 kg

Force F = 8 i + 2 j N

Magnitude of force F = | F | = F = [ 8² + 2²]½ =[64 + 4]½ = √68 = 2√17 N

Acceleration a in the body = F/m = (2√17 N)/0.5 kg = 4 √17 m/s²

Distance covered = avearge velocity × time interval = 4 m

=> ½ ( initial velocity + final velocity) × 1 second = 4 m

=> ½( 0 m/s + v m/s) × 1s = 4 m (because initially body is at resAt)

v (= velocity after 1second) = 8 m/s

But

v = u + a t = 0 + (4√17) × 1 = 4 × 4.4.12 × 1s = 16.48 m/s

We get two inconsistent results. The data given that distance covered in one second after starting from rest is 4 m gives velocity at end of 1 second as 8 m/s. But the data of force and mass which allows us to find acceleration gives for velocity at end of 1 second as 16.48 m/s. It is an inconsistent question.