A 5. 32-kg object passes through the origin at time t = 0 such that its x compon

Question

A 5. 32-kg object passes through the origin at time t = 0 such that its x component of velocity is 5. 45 m/s and its y component of velocity is -2. 73 m/s. What is the kinetic energy of the object at this time? J 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 N direction degree measured from the +x axis What is the speed of the particle at t = 2. 00 s? m/s A 7. 80-g bullet moving at 650 m/s penetrates a tree trunk to a depth of 4. 8 cm. Use work and energy considerations to find the average frictional force that stops the bullet. N Assuming the friction force is constant, determine how much time elapses between the moment the bullet enters the tree and the moment it stops moving.

Explanation / Answer

a)

Upper question:

m = 5.32 Kg

v0_x = 5.45 m/s

v0_y = -2.73 m/s

K = 0.5 m v^2

K = 0.5 m ((v0_x)^2 + (v0_y)^2)

K = 0.5 * 5.32 * (5.45*5.45 + 2.73*2.73)

K = 98.8 J

b)

x = 0.5 (a_x) t^2 + (v0_x) * t

==> 8.5 = 0.5 * a_x * 2*2 + 5.45*2

==> a_x = -1.2 m/s2

y = 0.5 (a_y) t^2 + (v0_y) * t

==> 5 = 0.5 * a_y * 2*2 - 2.73*2

==> a_y = 5.23 m/s2

magnitude = a = (1.2*1.2 + 5.23*5.23) = 5.37 m/s2

direction: = 180 + Arctan(5.23/1.2) = 257 degrees

c)

v_x = a_x t = -1.2 * 2 = -2.4 m/s

v_y = a_y t = 5.23 * 2 = 10.46 m/s

v = (2.4*2.4 + 10.46*10.46) = 10.7 m/s

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