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Ilya [14]
3 years ago
5

A 56.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.517 and 0.260,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed
Physics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

Explanation:

Magnitude of frictional force = μ mg

μ is either static or kinetic friction.

To start the crate moving , static friction is calculated .

a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .

force required =.517 x 56.6  x 9.8 = 286.76  N .

b ) to  slide the crate across the dock at a constant speed , force required

= μ mg where μ is coefficient of kinetic  friction , where μ is kinetic friction

= .26 x 56.6  x 9.8 = 144.21 N .

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Answer:

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Part 2) Maximum speed equals 39.66 m/s.

Explanation:

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v^{2}=u^2+2as

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Applying the given values we get

v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

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Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equalsT=2\times 30.51\approx61seconds

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