A machine in an ice factory is capable of exerting 3.00x10^2 N of force to pull a large block of ice up a slope. The block weighs 1.22 x 10^4 N. Assuming there is no friction, what is the maximum angle that the slop can make with the horizontal if the machine is able to complete the task?

14.24°

Explanation:

Force applied to the object up the plane is the moving force (Fm). The frictional force (Ff) is always acting up the plane. Since there is no friction, frictional force is 0N. Therefore the only force needed to move the body up the slope is the moving force. Mathematically,

Fm = Wsin (theta) {force acting along the plane}

Given Fm = 300N, Weight = 1220N

Substituting this values in the formula to get theta we have;

300 = 1220sin (theta)

sintheta = 300/1220

Sintheta = 0.246

theta = arcsin 0.246

theta = 14.24°

Therefore, the maximum angle that the slop can make with the horizontal if the machine is able to complete the task is 14.24°

1.41 °

Explanation:

From machine,

The formula for the efficiency of machine is

E (%) = (M. A/V. R) * 100 ... Equation 1

Where E (%) = Percentage Efficiency of the machine, M. A = Mechanical Advantage of the machine, V. R = Velocity ratio of the machine.

Note: A slope forms an an inclined plane with the horizontal

Make V. R the subject of the equation

V. R = 100M. R/E (%) ... Equation 2

But,

M. A = L/E ... Equation 3

Where Load, E = Effort.

Given: E = 3.00*10² N, L = 1.22*10⁴ N

Substitute into equation 3

M. A = 1.22*10⁴ / 3.00*10²

M. A = 40.7.

Note: Assuming no friction means that the efficiency of the inclined plane (slope) = 100 %

Hence, E (%) = 100 %

Substitute into equation 2

V. R = 100 (40.7) / 100

V. R = 40.7.

The velocity of an inclined plane is

V. R = 1/sin∅ ... Equation 4

Where ∅ = maximum angle that the slope makes with the horizontal.

make ∅ the subject of the equation

∅ = sin⁻¹ (1/V. R) ... Equation 5

Substitute the value of V. R into equation 5

∅ = sin⁻¹ (1/40.7)

∅ = sin⁻¹ (0.0246)

∅ = 1.41 °