A warehouse crate slides at a steady 1.2 m/s behind Casper, who keeps a horizontal force on the tow strap.
If he applied more force than the friction, the crate would speed up. It does not speed up. So how big is his force?
Constant velocity means equilibrium: zero net force on the crate.
Perpendicular to the floor, the weight and the normal force balance, and neither has any horizontal part. So the horizontal line decides everything here.
On the horizontal line that leaves two forces: the applied force forward and the kinetic friction backward.
So the steady-speed applied force equals the kinetic friction. Not more, not less.
June's box slides at constant velocity while she applies a 52 N horizontal force to it. June says the kinetic friction on the box is also 52 N. Is June correct?
If the friction is not given, compute it first: f_k = μ_k N, with N from the perpendicular balance.
Casper's crate slides at constant velocity. The surface pair has μ_k = 0.30, and the normal force on the crate is 220 N. Calculate the applied force that keeps the crate at constant velocity.
Write down the values in the question:
μ_k = 0.30
N = 220 N
applied force = ?
Find the kinetic friction:
f_k = μ_k N
f_k = 0.30 × 220
f_k = 66 N
Constant velocity means the applied force balances the friction:
applied force = f_k
The applied force is 66 N, exactly matching the kinetic friction.
A trunk slides at constant velocity. μ_k = 0.25 and N = 280 N. Enter the applied force that keeps it at constant velocity, in N.
A bigger applied force would mean a net forward force and a speeding-up crate. The equality is what steady means in force terms.
Gideon says a crate sliding at constant velocity needs an applied force larger than the kinetic friction. Is Gideon correct?
A sled slides at constant velocity while the kinetic friction on it is 48 N. Calculate the applied force on the sled.
Write down the values in the question:
f_k = 48 N
applied force = ?
Constant velocity means zero net force, so the applied force balances the friction:
applied force = f_k
The applied force is 48 N.
A pallet slides at constant velocity. μ_k = 0.35 and N = 300 N. Enter the applied force that keeps it at constant velocity, in N.
A hay bale slides at constant velocity. μ_k = 0.20 and N = 430 N. Enter the applied force that keeps it at constant velocity, in N.
A 20 kg feed sack slides at constant velocity over a level barn floor. μ_k = 0.30 and nothing presses on the sack. Take g = 9.8 N/kg. Enter the applied force that keeps it at constant velocity, in N.