101k views
21 votes
On earth, when a box slides across a horizontal board, the board exerts a frictional force of magnitude f0 on the box. The board and the box are moved to a planet with twice the radius but one-third the mass of earth. When the box slides across the board, the frictional force exerted by the board on the box is now.

1 Answer

6 votes

Answer:


(1/12)\, f_(0).

Step-by-step explanation:

Fact: The weight of the box on the said planet would be
(1/12) the weight on the earth:

Let
r and
M denote the radius and mass of planet earth, respectively. Let
m denote the mass of the box. Let
G denote the gravitational constant.

If the earth is a uniform sphere, the weight of this box on earth would be:


\begin{aligned}(G\, M\, m)/(r^(2))\end{aligned}.

On a spherical planet with mass
(M/3) and radius
2\, r, the weight of this box on that planet would be:


\begin{aligned}(G\, (M/3)\, m)/((2\, r)^(2)) &= (1)/(12)* (G\, M\, m)/(r^(2))\end{aligned}.

In other words, the weight of this box on the said planet would be
(1/12) of the weight of this box on the earth.

The board exerts (kinetic) friction on the box while the box slides across this board. The magnitude of this friction force is proportional to the magnitude of the normal force that the board exerts on the box.

Since this board is horizontal, the magnitude of this normal force would be equal to the weight of the box. Thus, the magnitude of the friction in this question would be proportional to the the weight of this box.

When the weight of this box is reduced (in magnitude) to
(1/12) of the value on the earth, the magnitude of this friction would also be reduced to
(1/12)\! of the original value. Thus, the magnitude of the friction on the said planet would be
(1/12)\, f_(0).

User Moh Tarvirdi
by
4.3k points