## c0c0pIe 3 years ago A 200 g particle gains a momentum of 25 Ns in 12 s starting from rest. Find the kinetic energy of the particle and the applied force.

We are given the mass of the particle m=200 g = 0.2 kg. And the momentum p = 25 Ns = 25 kg m/s. We know $$p=mv$$ and kinetic energy $$E_k = \frac{1}{2}mv^2$$. Multiplying numerator and denominator of the expression for $$E_k$$ by m, we get: $E_k = \frac{p^2}{2m}$ Plugging in our given quantities, yields: $E_k = 1562.5\ \mathrm{J}$ Assuming a constant force, we can use the impulse/momentum equation: $F\Delta t = \Delta p$ Since we are starting from rest $$\Delta p = p$$, and we are given $$\Delta t = 12\mathrm{s}$$. So that: $F = \frac{p}{\Delta t} = \frac{25\ \mathrm{Ns}}{12s} = 2.08\ \mathrm{N}$