Question

A 49 N box is to be pulled up a 10 degree ramp at an angle of 40 degrees to the ramp that is 5 m long. The box starts at the base of the ramp. Of the coef of Kinetic friction is 0.1, how hard do you have to pull to make the box to move up the ramp at a constant velocity?

Answer 1

Answer:

17.4 N

Explanation:

According to Newton's second law, the net force acting on the box along the ramp is equal to the product of mass and acceleration:

$F_{net}=ma$

If we want to move the box at constant velocity, this means acceleration equal to zero: a = 0, so the previous equation becomes:

$F_{net}=0$ (1)

Let's now understand what are the forces exerted on the box along the surface of the ramp:

- The pull (the applied force), let's call it F, which pulls up along the ramp. However, this pull is applied at an angle of 40 degrees, so the actual pull along the ramp will be $F cos 40^{\circ}$

- The component of the weight parallel to the surface of the ramp, equal to

$-mg sin \theta$

where (mg) = 49 N is the weight of the box and $\theta=10^{\circ}$ is the angle of the ramp. This force pulls down along the ramp, so we put a negative sign in front of it.

- The frictional force, equals to

$-\mu mg cos \theta$

where $\mu=0.1$ is the coefficient of friction. This force pulls down along the ramp (in the opposite direction of the motion), so we put a negative sign as well

Therefore, the equation for the net force (1) becomes:

$F cos 40^{\circ}-mg sin \theta -\mu mg cos \theta =0$

from which we can find the value of F:

$F=\frac{mg sin \theta + \mu mg cos \theta}{cos 40^{\circ}}=\frac{(49 N)(sin 10^{\circ})+(0.1)(49 N)(cos 10^{\circ})}{cos 40^{\circ}}=17.4 N$