Sally is working at the Ekka, and has to push a 10.0kg box of showbags up a 30 ramp to load them onto a truck.
The ramp is 3.0m long and we can assume that Sally applies a constant force to the box parallel to the ramp. The
box starts from rest.
a) If Sally wants to get the box up the ramp in 5.0s, what acceleration must it have?
b) What is the nett force on the box if it is accelerating at this rate?
c) If the coefficient of kinetic friction between the box and the ramp is 0.42, with what force must Sally push on the
box?
The ramp is 3.0m long and we can assume that Sally applies a constant force to the box parallel to the ramp. The
box starts from rest.
a) If Sally wants to get the box up the ramp in 5.0s, what acceleration must it have?
b) What is the nett force on the box if it is accelerating at this rate?
c) If the coefficient of kinetic friction between the box and the ramp is 0.42, with what force must Sally push on the
box?
-
a) In order to move three meters in 5.0s, you would need an acceleration of:
We use this kinematic equation:
x = xₒ + vₒ t + ½ a t²
Plugging in the values:
3.0m = 0 + 0*5.0 + ½ * a * 5.0s²
Solving for a:
3.0m = ½ * 25.0 * a = 12.5 * a
a = 3.0/12.5 = 0.24m/s²
b) The net force on a 10.0kg object that is accelerating at 0.24m/s² is:
F=ma
F=10.0 * 0.24
F= 2.4N
c) There are three forces acting on the box: Sally's push, weight, and friction.
Sally's push - (weight + friction) = The net force, which we found in the last problem.
We don't need the full weight, we need the amount of weight that is parallel to the surface of the ramp:
Weight = m * g * sin30 = 10.0 * 9.81 * sin30 = 49.05N
The force of friction is equal to the coefficient of kinetic friction times the normal force, and the normal force is equal in magnitude to the amount of weight that is perpendicular to the ramp.
Friction = coefficient * m * g * cos30 = 0.42 * 10.0 * 9.81 * cos30 = 35.68N
So:
Sally's push - (weight + friction) = The net force
Sally's push - (49.05N + 35.68N) = 2.4N
Sally's push - 84.73N = 2.4N
Sally's push = 87.13N
Done!
We use this kinematic equation:
x = xₒ + vₒ t + ½ a t²
Plugging in the values:
3.0m = 0 + 0*5.0 + ½ * a * 5.0s²
Solving for a:
3.0m = ½ * 25.0 * a = 12.5 * a
a = 3.0/12.5 = 0.24m/s²
b) The net force on a 10.0kg object that is accelerating at 0.24m/s² is:
F=ma
F=10.0 * 0.24
F= 2.4N
c) There are three forces acting on the box: Sally's push, weight, and friction.
Sally's push - (weight + friction) = The net force, which we found in the last problem.
We don't need the full weight, we need the amount of weight that is parallel to the surface of the ramp:
Weight = m * g * sin30 = 10.0 * 9.81 * sin30 = 49.05N
The force of friction is equal to the coefficient of kinetic friction times the normal force, and the normal force is equal in magnitude to the amount of weight that is perpendicular to the ramp.
Friction = coefficient * m * g * cos30 = 0.42 * 10.0 * 9.81 * cos30 = 35.68N
So:
Sally's push - (weight + friction) = The net force
Sally's push - (49.05N + 35.68N) = 2.4N
Sally's push - 84.73N = 2.4N
Sally's push = 87.13N
Done!