Question

An inclined plane of length 10m has a height of 6m. If the coefficient of friction between the block and the plane is 0.4. a. What force parallel to the plane is needed to slide the 50 kg block up the plane with a uniform speed ? b.What is the efficiency of the plane

Answer 1

Answer:

F = 489.93 N

Explanation:

I attached an image of the system below.

By using a rotated system of coordinates you have the following equation:

$F-F_f-W_x=ma$  (1)

F: force

Ff: friction force

Wx: x component of the weight = Wxsin(θ)

In this case you have that the speed of the block is constant, then a= 0m/s^2.

You also consider that the friction force is given by:

Wx = μNsin(θ) = μmgsin(θ)

μ: friction coefficient = 0.4

m: mass of the block = 50kg

g: gravitaional acceleration = 9.8m/s^2

θ is the angle of inclination of the inclined plane, you calculate this by using the following:

$sin^{-1}(\frac{6}{10})=36.86\°$

You replace the values of the parameters in the equation (1) with a = 0m/s^2

$F-F_f-W_x=0\\\\F=F_f+W_x\\\\F=\mu mg + mg sin(\theta)=mg(\mu+sin\theta)\\\\F=(50kg)(9.8m/s^2)(0.4+sin(36.86\°))=489.93\ N$