A 25.0-kg box is released at the top of a frictionless ramp that is inclined at an angle of 35.0◦ with the horizontal. The box slides 1.50 m down the ramp before you catch it and bring it to a stop in 0.400 m. Assuming you pushed on the box with a constant force that was parallel to the ramp as you caught it, what was the magnitude of the force you exerted on the box to bring it to a stop?
Using Newton's second law
Fnet = ma
mgsin = ma
a = gsin
a = 9.8 * sin 35
a = 5.621 m/s2
Now, we can use kinematics to know the final velocity of block after sliding for 1.50 m
initial velocity (u) = 0
v2 - u2 = 2ad
v = sqrt (2 * 5.621 * 1.50)
v = 4.1 m/s
Now, we start applying force on it in order to stop it.
so, here we want its final velocity to be zero.
v2 - u2 = 2ad
0 - u2 = 2ad
a = -u2 / 2d
a = - 4.12 / 2 * 0.4
a = - 21.07 m/s2
do not worry about negative sign it just means that block is slowing down
so,
F = ma
F = 25 * 21.07
F = 526.97 N
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