A 5.22-kg object passes through the origin at time t = 0 such that its x compone

Question

A 5.22-kg object passes through the origin at time t = 0 such that its x component of velocity is 5.15 m/s and its y component of velocity is -2.91 m/s. (a) What is the kinetic energy of the object at this time? 91.3J (b) At a later time t = 2.00 s, the particle is located at x = 8.50 m and y = 5.00 m. What constant force acted on the object during this time interval? magnitude 28.6 N direction 99.6

Explanation / Answer

Here we have a case of constant acceleration by the force found in part b of the question horizontal acceleration = -0.91 m/s^2 vertical acceleration = 5.40 m/s^2 consider horizontal direction : distance = 8.50 m acceleration = -0.91 m/s^2 initial speed = 5.15 m/s v^2 = u^2 + 2as => v^2 = 5.15^2 - 2*0.91*8.5 => v = 3.32 m/s consider vertical direction : distance = 5 m acceleration = 5.4 m/s^2 initial speed = -2.91 m/s v^2 = u^2 + 2as => v^2 = 2.91^2 + 2*5.4*5 => v = 7.9 m/s so the velocity = ( 3.32^2 + 7.9^2) ^1/2 = 8.57 m/s

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