Question

A box of mass 0.8 kg slide at speed of 10m/s.across a smooth level floor before it encounter a rough patch of length 3.0 m The friction force on the box due to this part of the floor is 70 N .What is the speed of box when it leave this rough surface ? What length of the rough surface would being the box completly to rest?

Answer 1

Answer on Question # 76349, Physics / Mechanics | Relativity

Question. A box of mass $0.8 \, kg$ slide at speed of $10 \, m/s$ across a smooth level floor before it encounter a rough patch of length $3.0 \, m$. The friction force on the box due to this part of the floor is $70 \, N$. What is the speed of box when it leaves this rough surface? What length of the rough surface would being the box completely to rest?

Given.

$$m = 0.8 \, kg; \, v = 10 \, m/s; \, F = 70 \, N; \, l = 3 \, m.$$

Find. $u, s$-?

Solution.

The kinetic energy of the box is

$$E = \frac{m v^2}{2} = \frac{0.8 \cdot 10^2}{2} = 40 \, J$$

The work done by friction is

$$A = F l = 70 \cdot 3 = 210 \, J$$

So, in this case, the box will not overcome 3 meters of the path and stop faster.

The box will stop at the point

$$E = F s \rightarrow s = \frac{E}{F} = \frac{40}{70} = 0.57 \, m$$

Similar results can be obtained as follows

$$a = \frac{F}{m} = \frac{70}{0.8} = 87.5 \, m/s^2$$ $$u = v - a t \rightarrow t = \frac{(v - u)}{a} = \frac{10 - 0}{87.5} = 0.11 \, s$$

Finally

$$s = v t - \frac{a t^2}{2} = 10 \cdot 0.11 - \frac{87.5 \cdot 0.11^2}{2} = 0.57 \, m$$

Answer. The box will not overcome 3 meters of the path and stop faster; $s = 0.57 \, m$.

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