Question

a 6kg cart on a level surgacr is pulled at a constant velocity of 2m/s by a constant force of 10 N. What is the acceleration ? What is the friction force opposing the motion?

Answer 1

A 6kg cart on a level surgacr is pulled at a constant velocity of $2\mathrm{m/s}$ by a constant force of $10\mathrm{N}$. What is the acceleration? What is the friction force opposing the motion?

Acceleration by definition equals:

$$a = \frac{dv}{dt}$$

if cart is pulled at a constant velocity $v = \text{const}$, then $\frac{dv}{dt} = 0$ and

$$a = 0$$

Newton's first law of motion: if $a = 0 \Rightarrow \sum \overrightarrow{F_t} = 0$, therefore $\vec{F} + \overrightarrow{F_{fr}} = 0$.

The force of friction directed opposite to motion, so $F - F_{fr} = 0$

$$F_{fr} = F = 10\ \mathrm{N}$$

Answer: $a = 0, F_{fr} = 10\ \mathrm{N}$