A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10m/s2? (c) Suppose you were per-forming the same experiment on the moon, where the acceleration due to gravity is 1.62m/s2. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

(a) Coefficient of static friction = 0.709

    Coefficient of kinetic friction = 0.471

(b) Force required = 257.5 N

(c) (i) Force required = 51.62 N.

(c) (ii) Acceleration = 3.86 m/s^2

Explanation:

Normal force of the crate = 45 * 9.81 = 441.45 N

Force required to overcome initial friction = 313 N

This would also mean that static frictional force equals 313 N.

Force required to move the crate at constant speed = 208 N

This means that the force of kinetic friction is also 208 N.

(a) Since we know both the normal force, and the static and kinetic frictional forces we can calculate the coefficients of friction as follows:

Static friction = Coefficient of static friction * Normal Force

313 = Coefficient of static friction * 441.45

Coefficient of static friction = 0.709

Kinetic friction = Coefficient of kinetic friction * Normal Force

208 = Coefficient of kinetic friction * 441.45

Coefficient of kinetic friction = 0.471

(b) Since force = mass * acceleration we can calculate the force required as follows:

Force required = 45 * 1.1

Force required = 49.5 N

This would be required ON TOP of the force required to overcome kinetic friction. So the total force required would be:

Force required = 49.5 + 208 = 257.5 N

(c) (i) Since the gravity is less as compared to earth, the normal force will be:

Normal force = mass * gravity

Normal force = 45 * 1.62

Normal force = 72.9

Keeping the same static friction coefficient as (a) we get:

Kinetic friction = 0.709 * 72.9

Kinetic friction = 51.69 N

Thus, the push required for it to move will be 51.62 N.

(c) (ii) Since we know the new normal force, the kinetic friction will be:

Kinetic friction = 0.471 * 72.9

Kinetic friction = 34.34 N

We can now calculate acceleration using the total force as follows:

Total force = mass * acceleration

208 - 34.34 = 45 * acceleration

Acceleration = 3.86 m/s^2

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

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