Given: The horizontal force moves the 3 kg package up a smooth hill whose profil

Question

Given:           

The horizontal force moves the 3 kg package up a smooth hill whose profile is given by y = x2/4, where x and y are in meters. It is known that the horizontal component of the velocity of the package is constant at 5 m/s. Friction is negligible.

Find:                       

a) Determine the velocity vector of the package in Cartesian coordinates when x = 3m.

b) Determine the acceleration vector of the package in Cartesian coordinates when x = 3m.

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Explanation / Answer

(a) Horizontal velocity, Vx = dx/dt = 5 m/s.
Vertical velocity, Vy = dy/dt = 2x/4 = x/2.

At x = 3 m, Vy = 3/2 = 1.5 m/s.

Hence, Velocity vector at x = 3 m, is given by 5i + 1.5 j where i = unit vector in x - direction and j = unit vector in y-direction.

(b) Horizontal Acceleration, ax = dVx/dt = 0.

Vertical Acceleration, ay = dVy/dt = 1/2 = 0.5.

Hence, acceleration vector at x = 3 m is given by, 0*i + 0.5*j = 0.5j where i = unit vector in x - direction and j = unit vector in y-direction.

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