A single force acts on a 7.0 kg particle-like object in such a way that the posi

Question

A single force acts on a 7.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 5.0 s. In the figure below, a 15 g bullet moving directly upward at 1000 m/s strikes and passes through the center of mass of a 3.0 kg block initially at rest. The bullet emerges from the block moving directly upward at 200 m/s. To what maximum height does the block then rise above its initial position? m

Explanation / Answer

Pblm 1

Work = change in Kinetic Energy
KE = (1/2)mv^2
Work = (1/2)*m*[v(final)^2 - v(initial)^2)]

v = d/dt x = d/dt (3t - 4t^2+ 1t^3)
v = 3 - 8t + 3t^2
v(final) = 3 - 8(5) + 3(5)^2 = 38 m/s
v(initial) = 3 - 8(0) + 3(0)^2 = 3 m/s

Work = (1/2)*7*[(38)^2-(3)^2]

           = 5022.5 J

Pblm 2

Conservation of momentum vertically,

0.015 * 1000 + 3 * 0 = 0.015 * 200 + 3 * u

u = 4 m/s

v=0; a= - 9.81

v2 = u2 + 2as
0 = 4^2 - 2 * 9.81 * s

s = 0.815 m -> answer

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