Question

A 500kg horse pulls a cart of mass 1500kg along a level road with an acceleration of 1m/s2. If the coefficient of sliding friction is 0.2 then the horizontal force exerted by the earth on the horse is
A. 300 kg-wt
B. 400 kg-wt
C. 500 kg-wt
D. 600 kg-wt

Answer 1

Hint – To give the answer of this question, we use the basic formula of friction i.e. f=$\mu N$; where$\mu$is coefficient of friction and N is the normal force between surface and object. Firstly we see the definition of friction. In laws of motion we study about friction and its properties. Also N= mg; where m is mass of object and g is acceleration due to gravity i.e. 9.8 m/$s^2$,here we use g=10m/$s^2$.

Complete answer:
Friction is a force between two surfaces that are sliding, or trying to slide, across each other. This force acts between two touching surfaces.
In this question, it is given that;
Mass of caw =1500kg, mass of horse =500kg, acceleration =1m/s2, and coefficient of friction i.e.$\mu$=0.2
So, N
Force on caw= force on caw by friction + force on caw by acceleration
=f +F
=$\mu mg$+ma F
=$0.2 \times 1500 \times 10$+$1500 \times 1$
=4500 $\mu$
=450g
And,
Force on horse = force on horse by friction + force on horse by acceleration
= f +F
=$\mu mg$+ma
=$0.2 \times 500 \times 10$+$500 \times 1$
=1500
= 150g
Thus, Total force =450g +150g
=600g
=600kg-wt
Hence, Option D is the correct option.

Note – Here we use F=ma; where f= force on body, m= mass of body, and a= acceleration of body. Frictions have two types’ first static friction and second kinetic friction. Kinetic friction acts when an object in motion and static friction acts on two surfaces that are not moving relatively, and the value of static friction is more than kinetic friction.