An object of mass 0.500 kg is released from the top of a building of height 7.00

Question

An object of mass 0.500 kg is released from the top of a building of height 7.00 m. The object experiences a horizontal constant force of 1.20 N due to a wind blowing parallel to the face of the building. Draw a free-body diagram of the forces acting on the object. Applying Newton's Laws of motion and the kinematic equations, find Find the time it takes for the object to strike the ground. What is the magnitude of the acceleration of the object? Through what horizontal distance does the object move before it hits the ground?

Explanation / Answer

a) There is normal force on the body.

so, remove upwrd arrow.

draw downward arrow for garvity.

draw an arrow towards right for air resistance force.

b) u = 0 (initial speed)

h = 7 m

Apply, h = u*t + 0.5*g*t^2

7 = 0 + 0.5*9.8*t^2

==> t = sqrt(7/(0.5*9.8))

= 1.195 s <<<<<<<<<<<<----------------Answer

c) ax = Fx/m = 1.2/0.5 = 2.4 m/s^2

ay = g = 9.8 m/s^2

so, a = sqrt(ax^2 + ay^2)

= sqrt(2.4^2 + 9.8^2)

= 10.09 m/s^2 <<<<<<<<<<<<----------------Answer

d)x = ux*t + 0.5*ax*t^2

= 0 + 0.5*2.4*1.195^2

= 1.71 m <<<<<<<<<<<<----------------Answer

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