Question

question 1
adjust the applied force on the crate until the velocity is constant at 5 m/s.
(hint: youll need to use a big force to start the crate moving, but then decrease it to go a constant velocity.
if the speed changes at all, keep adjusting!)
a what is the applied force? 0 n
b. what is the friction force? select n
c. what is the net force? select n

Explanation:

Step1: Recall Newton's first - law

When an object moves with a constant velocity, the net force acting on it is zero according to Newton's first law ($F_{net}=0$).

Step2: Analyze the forces acting on the crate

The forces acting on the crate are the applied force $F_{applied}$ and the friction force $F_{friction}$. The net force $F_{net}=F_{applied}-F_{friction}$. Since $F_{net} = 0$ (constant velocity), $F_{applied}=F_{friction}$.

Step3: Determine the net - force

As the crate is moving at a constant velocity, the acceleration $a = 0$. According to Newton's second law $F_{net}=ma$, and since $a = 0$, $F_{net}=0$ N.

Answer:

A. The applied force: The value of the applied force is equal to the friction force when the crate is moving at a constant velocity. Without knowing the coefficient of friction and the normal force, we can't give a numerical value, but when the crate is moving at constant velocity, if we assume the applied force found through adjustment is $F$, then the answer is $F$ N.
B. The friction force: Since $F_{net}=F_{applied}-F_{friction}=0$ for constant - velocity motion, the friction force is equal to the applied force. So it is the same non - zero value as the applied force (when the crate is moving at constant velocity), say $F$ N.
C. The net force: According to Newton's first law for an object moving with constant velocity, the net force $F_{net}=0$ N.