Question

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?

Answer 1

Using Newton's second law

F_net = 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