Question

A child exerts a force on an m = 35 kg box that is parallel to an incline where = 30° so that the box moves up the plane at constant speed. The coefficient of kinetic friction between the box and the plane is 0.24.(a) Calculate the normal force on the box.(b) Calculate the component of the box's weight force that is parallel to the plane.(c) Calculate the frictional force acting on the box.(d) Calculate the force applied by the student to the box.

Answer 1

Given,

The mass of the box, m=35 kg

The angle of inclination, θ=30°

The coefficient of kinetic friction between the box and the plane. μ=0.24

(a)

The normal force on the box is equal to the vertical component of the weight of the box, which is given by,

`N=mg\cos \theta`

Where g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} N=35*9.8*\cos 30^(\circ) \\ =297.05\text{ N} \end{gathered}`

Thus the normal force on the box is 297.05 N

(b)

The component of the box's weight that is parallel to the plane is given by,

`W_x=mg\sin \theta`

On substituting the known values,

`\begin{gathered} W_x=35*9.8*\sin 30^(\circ) \\ =171.5\text{ N} \end{gathered}`

Thus the vertical component of the weight of the box is 171.5 N

(c)

The frictional force is given by,

`f=N\mu`

On substituting the known values,

`\begin{gathered} f=297.05*0.24 \\ =71.29\text{ N} \end{gathered}`

Thus the frictional force acting on the box is 71.29 N

(d)

As the box is moving at a constant speed, the net horizontal force on the box is zero. Therefore the force applied by the student must be equal to the sum of the frictional force and the horizontal component of the weight of the box.

Therefore, the force applied by the student must be equal to

`F=f+W_x`

On substituting the known values in the above equation,

`\begin{gathered} F=71.29+171.5 \\ =241.79\text{ N} \end{gathered}`

Thus the force applied by the student is 241.79 N