We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30 o . A worker,
ID: 1472191 • Letter: W
Question
We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30 o. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But the friction is NOT negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. A) Find the magnitude of the friction force acting on the crate, assuming that it is constant. B) How fast is the crate moving when it reaches the bottom of the ramp? We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30 o. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But the friction is NOT negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. A) Find the magnitude of the friction force acting on the crate, assuming that it is constant. B) How fast is the crate moving when it reaches the bottom of the ramp? We want to slide a 12 kg crate up a 2.5 m long ramp inclined at 30 o. A worker, ignoring friction, calculates that he can do this by giving it an initial speed of 5.0 m/s at the bottom and letting it go. But the friction is NOT negligible; the crate slides only 1.6 m up the ramp, stops, and slides back down. A) Find the magnitude of the friction force acting on the crate, assuming that it is constant. B) How fast is the crate moving when it reaches the bottom of the ramp?Explanation / Answer
A)
(initial kinetic energy) = (gravitational potential energy when crate stops) + (work done by friction)
(1/2)mv² = mgh + Fz
(1/2)(12 kg)(5 m/s)² = (12 kg)(9.81 m/s²)(1.6 m)(sin30°) + (F)(1.6 m)
F = 34.9 N
B)
(gravitational potential energy when crate is stopped) = (final kinetic energy) + (work done by friction)
mgh = (1/2)mv² + Fz
(12 kg)(9.81 m/s²)(1.6 m)(sin30°) = (1/2)(12 kg)v² + (34.9 N)(1.6 m)
v = 2.53 m/s