00:02
In this problem, a person is pulling a cart, okay, and the acceleration of the person is, let's say, a, and which is 0 .1 meter per second square.
00:17
Since the rope is inextensible in elastic, therefore the acceleration of the cart will be same as the acceleration of the person.
00:26
Okay, now let's draw the free ball diagram of the cart.
00:30
So this is the cart.
00:32
The goal is inside.
00:35
Now the forces acting on the cart is the frictional force acting opposite to the direction of motion and the tension along the direction of motion.
00:46
So from newton's second law of motion, the summation of all forces along the direction of motion is equivalent to the mass times the acceleration.
00:55
Or we can say t minus the frictional force is equivalent to the mass.
01:01
M1 times its acceleration or the tension in the rope is equivalent to m1a plus the fictional force now for part a let's substitute the value to find the tension in the rope so t is equivalent to the mass of the cart and the girl combined is 35 kilograms so 35 kilograms into the acceleration and that is 0 .1 meter per second square plus the fictional force and that is 30 so from this the tension in the rope is equivalent to 33 .5 newtons.
01:41
Okay, now part b.
01:43
Now let's consider the person pulling the cart, the free border diagram...