A man pushed a truck initially at rest and weighting $5 t o n s$ along a horizontal rail with a steady force of $400 N$ . the resistance to motion amount to $60 N / t o n$ . The velocity of the truck at the end of $30$ seconds is:

6 m / s

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0.6 m / s

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3 m / s

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0.3 m / s

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Solution

The correct option is B $0.6 m / s$
Given,
Initial velocity, $u = 0$
Mass of truck, $M = 5 t o n = 5000 k g$
Resistive force per unit ton, $μ = 60 N / t o n$
Forward force, $F_{f} = 400 N$
Resisting force, $F_{r} = μ M = 60 \times 5 = 300 N$
$F_{n e t} = F_{f} - F_{r}$
$M a = 400 - 300$
$a = \frac{100}{5000} = 0.02 m s^{- 2}$
Apply kinematic equation of motion
$v = u + a t$
$v = 0.02 \times 30 = 0.6 m s^{- 1}$
Hence, velocity of truck after $30 s e c$ is $0.6 m s^{- 2}$ .

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