Question

The student in item 1 moves the box up a ramp inclined at 12° with the horizontal. if the box starts from rest at the bottom of the ramp and is pulled at an angle of 25.0° with respect to the incline and with the same 185 n force, what is the acceleration up the ramp? assume the coefficient of kinetic friction is 0.27.

Answer 1

To find the acceleration up the ramp, resolve the force applied into two components: one parallel to the ramp and one perpendicular to the ramp. The force parallel to the ramp can be found by multiplying the applied force by the sine of the angle between the force and the ramp. The net force parallel to the ramp will be the force parallel to the ramp minus the force of friction.

Step-by-step explanation:

To find the acceleration up the ramp, we need to resolve the force applied into two components: one parallel to the ramp and one perpendicular to the ramp. The component parallel to the ramp will contribute to the acceleration. The force parallel to the ramp can be found by multiplying the applied force by the sine of the angle between the force and the ramp. So, the force parallel to the ramp is 185 N × sin(25°) = 77.9 N.

The net force parallel to the ramp will be the force parallel to the ramp minus the force of friction. The force of friction can be found by multiplying the coefficient of kinetic friction (0.27) by the normal force, which can be found by multiplying the weight of the box by the cosine of the angle of the ramp. So, the force of friction is 0.27 × (2 kg × 9.8 m/s^2 × cos(12°)) = 5.63 N.

The net force parallel to the ramp is 77.9 N - 5.63 N = 72.27 N. To find the acceleration, we divide the net force parallel to the ramp by the mass of the box. So, the acceleration up the ramp is 72.27 N / 2 kg = 36.14 m/s^2.

Answer 2

Answer: Fg,y = mg(cos q1) Fg,x = mg(sin q1) Fapplied,x = Fapplied(cos q2) Fapplied,y = Fapplied(sin q2) Fn = Fg,y â’ Fapplied,y Fk =mkFn Fapplied,x â’ Fk â’ Fg,x ax=fapplied-fk-fg,x/ m