A 0.5 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As

Question

A 0.5 kg block of ice is sliding by you on a very slippery floor at 2.5 m/s. As it goes by, you give it a kick perpendicular to its path. Your foot is in contact with the ice block for 0.003 seconds. The block eventually slides at an angle of 26 degrees from its original direction (labeled ? in the diagram). The overhead view shown in the diagram is approximately to scale. The arrow represents the average force your toe applies briefly to the block of ice.

Please show your work. I really am trying to learn.

Explanation / Answer

1) B

Because, direction of block changes at the instant force is applied, after the force is removed it moves in a staright line)

2) z component

since the force is applied in z direction.

3) tan(theta) = vz/vx

vz = vx*(tan(theta)

= 2.5*tan(26)

= 1.22 m/s

p = m*vx i - m*vz k

= 0.5*2.5 i - 0.5*1.22 k

= (1.25 i - 0.61 k )kg.m/s

4) Pfx = Pix i

= m*vx i

= 0.5*2.5 i

= 1.25 i kg.m/s

5) |P| = sqrt(Px^2 + Pz^2)

= sqrt(1.25^2 + 0.61^2)

= 1.391 kg.m/s

6) pfz = -0.61 k kg.m/s

7) Favg = delta_Pz/t

= ( -0.61 - 0)/0.003

= -203.3 k N

|Favg| = 203.3 N

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