A 25.0 kg block is initially at rest on a horizontal surface. A horizontal force of 75.0 N is required to set the block in motion, after which a horizontal force of 60.0 N is required to keep the block moving with constant speed. Find (a) the coefficient of static friction.
(b) the coefficient of kinetic friction between the block and the surface.

(a) 0.31

(b) 0.245

Explanation:

(a)

F' = μ'mg Equation 1

Where F' = Horizontal Force required to set the block in motion, μ' = coefficient of static friction, m = mass of the block, g = acceleration due to gravity.

make μ' the subject of the equation above

μ' = F'/mg Equation 2

Given: F' = 75 N, m = 25 kg

constant: g = 9.8 m/s²

Substitute these values into equation 2

μ' = 75/(25×9.8)

μ' = 75/245

μ' = 0.31.

(b) Similarly,

F = μmg Equation 3

Where F = Horizontal force that is required to keep the block moving with constant speed, μ = coefficient of kinetic friction.

make μ the subject of the equation

μ = F/mg Equation 4

Given: F = 60 N, m = 25 kg, g = 9.8 m/s²

Substitute these values into equation 4

μ  = 60/(25×9.8)

μ = 60/245

μ = 0.245

The magnitude of the acceleration is 1.2 m/s2 and the direction is positive, upward.

Explanation:

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky driver, and, thus, slow him down. It means, if the drag force is greater than the weight, acceleration should be upward and positive to slows him.

We have the weight W = 915 N of the sky driver. This is negative, due to the direction.

The drag force D = 1027 N. This is positive, due to the direction upward. The mass of the sky driver is m=93.4 kg with an unknown acceleration a.

The Newton laws shoued us:

∑F = m*a

D - W = m*a

1027 - 915 = 93.4*a

93.4*a = 112

a = 112/93.4

a = 1.2 m/s2

Finally, we can conclude that The magnitud of the acceleration is 1.2 m/s2 and the direction is positive, upward.