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bezimeni [28]
2 years ago
11

A 49.6-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.584 and 0.399,

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:
NISA [10]2 years ago
3 0

Answer:

(a) Force must be grater than 283.87 N

(B) Force will be equal to 193.945 N      

Explanation:

We have given mass of the crate m = 49.6 kg

Acceleration due to gravity g=9.8m/sec^2

Coefficient of static friction \mu _s=0.584

Coefficient of kinetic friction \mu _k=0.399

(a) Static friction force is given by F_S=\mu _smg=0.584\times 49.6\times 9.8=283.8707N

So to just start the crate moving we have to apply more force than 283.87 N

(B) This force will be equal to kinetic friction force

We know that kinetic friction force is given by F_k=\mu _kmg=0.399\times 49.6\times 9.8=193.945N

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Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
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i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

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Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

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(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

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