Question

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newton directed up the slope.The coefficient of sliding friction is 0.18.What is the mass of the box?

Answer 1

Mass of the box= 31.12 kg

Step-by-step explanation:

Second Newton's Law

It can be understood as the fact that if we sum all the forces applied to an object, we get the net force which will cause it to change its velocity in the direction of the net force. If the net force is zero, then the object will continue to be as it was (constant speed or at rest).

The reciprocal statement is also valid, i.e. if an object is moving at constant velocity then we can be sure the net force on it is zero.

The box of books is being moved upwards at a constant speed, then we know all acting forces will sum cero or will be balanced. Let's assume the x-axis placed parallel to the ramp with the positive direction pointing up to where the object is moving. The figure below shows all the forces acting on the x-axis:

The applied force
`F_a` (positive), the x-component of the weight
`W_x` (negative), and the friction force
`F_r` (negative). Recall the friction force is always opposite to movement.

The net force on the box is

`F_n=F_a-W_x-F_r`

We already know that
`F_n` is zero because the box is moving at a constant velocity, so

`F_a-W_x-F_r=0\ \ ......[eq 1]`

We need to know the mass of the object, so we have to prepare all the magnitudes to find m. The weight of the box (pointed downwards) is

`W=mg`

The x-component of that weight can be found by the construction of the diagram of the triangle formed by the forces and the known angle
`\theta`

`W_x=mgsin\theta`

The friction force can be obtained from the y-component of the weight which is equal to the normal force exerted by the surface of the ramp

`F_r=\mu mgcos\theta`

Replacing all in eq 1

`F_a-mgsin\theta-\mu mgcos\theta=0`

Rearranging and factoring

`\displaystyle mg(sin\theta+\mu cos\theta)=F_a`

Solving for m:

`\displaystyle m=(F_a)/(g(sin\theta+\mu cos\theta))`

`F_a=200 N,\ \theta=30^o,\ \mu=0.18`

`\displaystyle m=(200)/(9.8(sin30^o+0.18 cos30^o))`

`m=31.12\ kg`