Question
Question: A\[500kg\]horse pulls a cart of mass\[1500kg\]along a level road with an acceleration of\[1m{{s}^{-2...
A500kghorse pulls a cart of mass1500kgalong a level road with an acceleration of1ms−2. If the coefficient of sliding friction is0.2, then the force exerted by the horse in forward direction is
A)4500NB)4000NC)5000ND)6000N
Solution
Hint : Here if the horse needs to move the cart with the given acceleration, it will have to overcome the frictional force exerted by earth on the weight of the cart. This force will be given by the product of coefficient of sliding friction to the weight of the cart. Now if it needs to acquire the given acceleration it will need to exert more force. So the total force will be the sum of both force needed to overcome friction and force needed to gain the acceleration.
Formula Used: Ffriction=μ×Mtotalg
F=Mtotala
Complete Solution: Firstly we will look at the frictional force exerted by the earth on the masses. The coefficient of sliding friction is given by0.2.
The combined mass of horse and cart is Mtotal=mhorse+mcart=500+1500=2000kg
Now, the frictional force due to sliding is given by, Ffriction=μ×Mcartg
Ffriction=0.2×1500×10=3000N
So, the force exerted by earth on the cart is 3000N.
Now, if we calculate the force needed for acquiring the acceleration of1ms−2. It will be,