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Vinvika [58]
3 years ago
14

Example: A wooden crate with mass 100kg is at rest on a stone floor. You know that the coefficients of kinetic and static fricti

on are: μs=0.5 and μk=0.4. A) What is the minimum horizontal force F need to just get the crate moving?
Physics
1 answer:
alexgriva [62]3 years ago
7 0

Answer

Any force greater 490N

Explanation

The force required just to make an object slide over a rough horizontal surface is any force greater that the static friction which given by;

F=\mu_s mg.............(1)

Given;

\mu_s=0.5\\m=100kg\\g=9.8m/s^2

Hence;

F = 0.5 x 100 x 9.8

F = 490N.

We will only need the coefficient of kinetic friction if we were asked to find the force required to keep the object moving uniformly. Usually, the force needed to keep an object moving uniformly over a rough surface is lesser that which is needed to start its motion.

In this problem, we were only asked to find the minimum force required to make the object move which we have done.

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Read 2 more answers
Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
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Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

\underset{V_{B}}{\rightarrow} = velocity of car B before collision = V_{B} i + 0 j

\underset{V_{AB}}{\rightarrow} = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

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M_{B} = mass of car B = 1450 kg

Using conservation of momentum

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(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

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