a) A projectile of mass m = 7.98 kg is fired straight upward twice, both times a

Question

a)

A projectile of mass m = 7.98 kg is fired straight upward twice, both times at initial speed vo = 68.2 m/s. The first time there is no air friction, and the projectile reaches height ho; the second time the air dissipates Elost = 4435 J of energy as the projectile rises, and it rises to height h. Find ho - h, the difference betwen the height the projectile reaches without friction and the height it reaches with friction.

b)A one-dimensional force acts on a particle of mass m = 2.06 kg in such a way that its position is given by:

Find W, the work done by this force during the first 4.95 s.

Explanation / Answer

a)mgha=.5*m*v^2 so ha=237.31 mghb=.5*m*v^2-4435 so hb=180.6 b)x = 0.256t3 - 75.3t dx/dt=.768 t^2-75.3 d^2x/dt^2=1.536 t so F=md2x/dt2=2.06*1.536 t so W=int(F.dx)=int(2.06*1.536 t * (.768 t^2-75.3) *dt)=int(3.16416 *(.768t^3-75.3 t))=-2554.7

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