A 500 kg horse pulls a cart of mass 1500 kg along a level horizontal road with an acceleration of 1m/s. If coefficients of the sliding friction between cart and ground is 0.2 then the force exerted by the horse on the cart in forward direction is:
f = 1500 * g * .2
To find the force exerted by the horse on the cart in the forward direction, we need to consider the forces acting on the system.
The force of gravity acts downward on both the horse and the cart. As the road is level, the normal force exerted by the ground counteracts the force of gravity and cancels it out.
The only other horizontal force acting on the system is the force of friction between the cart and the ground. According to the problem, the coefficient of sliding friction is 0.2.
The equation for the force of friction can be written as:
Friction = coefficient of friction × normal force
Since the normal force cancels out the force of gravity, we do not need to consider it in this calculation.
Thus, the force of friction can be written as:
Friction = coefficient of friction × force of gravity
The force of gravity can be calculated as the product of mass and acceleration due to gravity:
Force of gravity = mass × acceleration due to gravity
In this problem, the acceleration due to gravity is not given, but we can assume it to be approximately 9.8 m/s^2.
So, the force of gravity can be calculated as:
Force of gravity = (mass of the horse + mass of the cart) × acceleration due to gravity
Once we have the force of gravity, we can calculate the force of friction using the given coefficient of friction.
Finally, the force exerted by the horse on the cart in the forward direction is equal and opposite to the force of friction.