Question

A horizontal force of only 5 N moves a cake of ice with constant speed across a floor (µ=0.1).
(a) What is the weight of the ice?

Answer 1

Answer on Question#40150, Physics, Mechanics | Kinematics | Dynamics

Question:

A horizontal force of only $5\mathrm{N}$ moves a cake of ice with constant speed across a floor $(\mu = 0.1)$ .

(a) What is the weight of the ice?

Answer:

Newton's first law of motion:

$$x: \quad F = F _ {f r}$$ $$y: \quad N = W$$

Friction force equals $F_{fr} = \mu N = \mu W$ , $\mu$ - coefficient of friction.

Therefore:

$$\mu W = F$$ $$W = \frac {F}{\mu} = \frac {5}{0 . 1} N = 5 0 N$$

Answer: $50N$