A 14.0 kg block of ice is at rest on a frictionless horizontalsurface. A force o

Question

A 14.0 kg block of ice is at rest on a frictionless horizontalsurface. A force of 14.0N is applied in the +ve x direction for2.00 sec. a) What is the velocity of the block after 2.00 seconds? b) What is the linear momentum of the block after 2.00seconds? c) What is the kinetic energy of the block after 2.00seconds? A 14.0 kg block of ice is at rest on a frictionless horizontalsurface. A force of 14.0N is applied in the +ve x direction for2.00 sec. a) What is the velocity of the block after 2.00 seconds? b) What is the linear momentum of the block after 2.00seconds? c) What is the kinetic energy of the block after 2.00seconds?

Explanation / Answer

Initial velocity (u) of the block = 0 Mass (m) of the block = 14.0 kg Force (F) applied on the block = 14.0 N THen from the Newton's second law of motion we have               F = ma               a = F / m                  = (14.0N) / (14.0kg)                  = 1.0m/s2 TIme given is t = 2.0s (a) Then from the kinematic equations of motion we have               v = u + at                  =0 + (1.0m/s2)(2.0s)                  =2.0 m/s (b) Linear momentum after 2.0s is p = mv                           p = (14.0kg)(2.0m/s) = 28.0 kg.m/s (c) Kinetic energy after 2.0s is K = (1/2)mv2                                           =(1/2)(14.0kg)(2.0m/s)2                                          = 28.0 J

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