A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 310N. After that you must reduce your push to 231N to keep it moving at a steady 25.2cm/s.

Part A: What is the coefficient of static friction between the crate and the floor?

Part B: What is the coefficient of kinetic friction between the crate and the floor?

Part C: What push must you exert to give it an acceleration of 1.13m/s^2? (in N)

Part D: Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 m/s^2. What magnitude push would cause it to move? (in N)

Part E: What would its acceleration be if you maintained the push in part C?

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Question

Physics

Kamora Stanley

9 June, 22:35

A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 310N. After that you must reduce your push to 231N to keep it moving at a steady 25.2cm/s.

Part A: What is the coefficient of static friction between the crate and the floor?

Part B: What is the coefficient of kinetic friction between the crate and the floor?

Part C: What push must you exert to give it an acceleration of 1.13m/s^2? (in N)

Part D: Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 m/s^2. What magnitude push would cause it to move? (in N)

Part E: What would its acceleration be if you maintained the push in part C?

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Aurora Sherman · Answer 1

A)

A force of 310 N initiates motion in mass 45.2 kg so

Coefficient of static friction = force / normal reaction

= force / weight

= 310 / (45.2 x 9.8)

= 0.7

B)

A force of 231 N maintains uniform motion in mass 45.2 kg so

Coefficient of kinetic friction = force / normal reaction

= force / weight

= 231 / (45.2 x 9.8)

= 0.52

C)

In motion, friction force acting = 231 N.

If F be the required force

F - 231 = 45.2 x 1.13 (net force = mass x acceleration)

= 282.07 N

D)

On moon g = 1.62

Force of static friction there

= weight x coefficient of friction

= 45.2 x 1.62 x. 7

= 51.25 N

Magnitude of push required = 51.25 N.