Lotte applies a steady 40 N horizontal force to a 6.0 kg box sliding on a level floor. On a frictionless floor, that force alone would give the box about 6.7 m/s² of acceleration.
The box actually gains speed at about 3.7 m/s². Where did the rest go?
Kinetic friction acts backward on the sliding box. Only part of the applied force is left to accelerate it.
Finding the acceleration takes four short stages, and each one is a move you already own.
Stage 1: draw the free-body diagram. Applied force forward, kinetic friction backward, weight down, normal force up.
Stage 2: find the normal force. The box does not accelerate perpendicular to the floor, so the perpendicular forces balance. Nothing presses on Lotte's box and nothing lifts it, so here N = mg.
Stage 3: find the kinetic friction, f_k = μ_k N, and subtract it from the applied force. The two act in opposite directions, so the net force is the applied force minus f_k.
Stage 4: finish with a = F_net/m.
Lotte applies a 40 N horizontal force to a 6.0 kg box sliding on a level floor. μ_k = 0.30. Take g = 9.8 m/s². Calculate the box's acceleration.
Write down the values in the question:
m = 6.0 kg
applied force = 40 N
μ_k = 0.30
a = ?
Find the normal force from the perpendicular balance:
N = mg
N = 6.0 × 9.8
N = 58.8 N
Find the kinetic friction:
f_k = μ_k N
f_k = 0.30 × 58.8
f_k = 17.64 N
Find the net force:
F_net = 40 − 17.64
F_net = 22.36 N
Write down the equation:
a = F_net/m
Substitute in the values:
a = 22.36/6.0
Simplify:
a = 3.73 m/s²
The box accelerates at 3.73 m/s², which is about 3.7 m/s².
If something presses down or lifts up, go back to the perpendicular balance in stage 2 before touching f_k. The normal force is not always mg.
A 10 kg sled slides on a level path while a 65 N horizontal force is applied to it. μ_k = 0.20. Take g = 9.8 m/s². Calculate the sled's acceleration.
Write down the values in the question:
m = 10 kg
applied force = 65 N
μ_k = 0.20
a = ?
Find the normal force from the perpendicular balance:
N = mg
N = 10 × 9.8
N = 98 N
Find the kinetic friction:
f_k = μ_k N
f_k = 0.20 × 98
f_k = 19.6 N
Find the net force:
F_net = 65 − 19.6
F_net = 45.4 N
Write down the equation:
a = F_net/m
Substitute in the values:
a = 45.4/10
Simplify:
a = 4.54 m/s²
The sled accelerates at 4.54 m/s².
A 4.0 kg crate slides on a level floor. μ_k = 0.35. Take g = 9.8 m/s². Enter the kinetic friction force on the crate in N.
The same 4.0 kg crate: the applied force on it is 30 N and the kinetic friction is 13.7 N. Enter the net force on the crate in N.
A 5.0 kg tub slides on a level floor while a 44 N horizontal force is applied to it. μ_k = 0.30. Take g = 9.8 m/s². Enter the tub's acceleration in m/s².
A 10 kg cart slides on a level floor while a 75 N horizontal force is applied to it. μ_k = 0.30. Take g = 9.8 m/s². Enter the cart's acceleration in m/s².
A 2.0 kg skillet slides on a counter while a 25 N horizontal force is applied to it. μ_k = 0.40. Take g = 9.8 m/s². Enter the skillet's acceleration in m/s².
Warehouse staff drag a loaded 15 kg bin over a level floor with a 120 N horizontal force. The bin is sliding, and the surface pair has μ_k = 0.30. Take g = 9.8 m/s². Enter the bin's acceleration in m/s².