1.

The motion of a particle of mass 0.1 kg is described by y = \(0.3 t+\frac{9.8}{2} t^{2}\). Find out the force acting on the particle.

Answer»

Given; m = 0.1kg; y = \(0.3 t+\frac{9.8}{2} t^{2}\); F = ?
∴ Velocity v = \(\frac{d y}{d t}=\frac{d}{d t}\left(0.3 t+\frac{9.8}{2} \cdot t^{2}\right)\)
or v = 0.3 + \(\frac{9.8}{2}\) × 2t
or v = 0.3 + 9.8t
Now, the acceleration,
a = \(\frac{d v}{d t}=\frac{d}{d t}(0.3+9.8 t)\) = 0 + 9.8
or a = 9.8 m/ s2
Therefore, the force acting on the particle,
F = ma = 0.1 × 9.8
or F = 0.98 N


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