What Is the Box's Speed at t = 0.6?

[the_d]

A 6 kg box slides down a long, frictionless
incline of angle 34 degrees. It starts from rest at time
t = 0 at the top of the incline at a height 17 m
above the ground. The acceleration of gravity is 9.81 m/s^2 :Find the box's speed at t = 0.6.

to do this problem i used the conservation of mechanical energy formula and just solved for v #2 squared but apparently that wasnt right

 

[xanthym]

A 6 kg box slides down a long, frictionless
incline of angle 34 degrees. It starts from rest at time
t = 0 at the top of the incline at a height 17 m
above the ground. The acceleration of gravity is 9.81 m/s^2 :Find the box's speed at t = 0.6.

to do this problem i used the conservation of mechanical energy formula and just solved for v #2 squared but apparently that wasnt right

You can't directly use Conservation of Energy with the (17 meter) incline height because the box has NOT slid down to the bottom in (t = 0.6 sec). Instead, try resolving the force of gravity into components parallel and perpendicular to the slide, determine the box's acceleration "a_parallel" parallel to the incline, and then use the formula below at (t = 0.6 sec) with (v₀ = 0):
v_parallel(t) = v₀ + a_parallel*t


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[the_d]

thanx xanthym