00:01
Hello students, in this question we are given a 20 kggy tongue is pushed across the floor of a moving van by a horizontal force of 60 newtons.
00:09
So let's show the fpd for the trunk.
00:11
Let's say this is our trunk and here is the ground of the van and it is pushed by a force of 60 newtons.
00:20
So let's say here we apply a force of 60 neutens and its mass is given as 20 kgues.
00:26
So gravitational force is acting like this of 200 newton's, mg 200 newtons.
00:32
Okay and we also given the coefficient of kinetic friction between the trunk in the floor as mu equals to 0 .255 okay, so we can see that if kinetic friction the coefficient of kinetic friction is that that means kinetic friction is acting and it is acting in the opposite direction to which this trunk is moving so the trunk is moving in this direction so the kinetic friction is acting in this direction as so though we have to calculate the value of magnitude of the friction force in the first part so for the first part we'll write the value of friction force as f will be equal to mu m g where mu is the coefficient of kinetic friction and mj is the weight okay so the value of mu is 0 .255 times the value of mass is 20 and the value of g is 10 so this will be equal to 51 neutins so, 51 newton's is acting on this trunk as prictional force.
01:38
So in the b part, we have to calculate the acceleration of the trunk.
01:42
Okay.
01:43
So to calculate the acceleration of the trunk, we can write it as force is equal to mass into mass times acceleration.
01:52
According to the newton's first look.
01:55
So here we will use the forces f net...